Determine the x-coordinates of the local
maxima and minima (if any) and intervals where f(x) is increasing or decreasing.
Determine intervals where f(x) is concave
upwards or downwards, and the x coordinates of inflection points (if any).
You may use, without verifying it, the formula
f′′(x)=4(3−x)3/2(3x−12).
There is a point at which the tangent line to the
curve y=f(x) is vertical. Find this point.
Sketch the graph y=f(x), showing
the features given in items (a) to (d) above and giving the (x,y) coordinates
for all points occurring above.
Hint+
You'll find the intervals of increase and decrease. These will give you a basic outline of the behaviour of the function. Use concavity to refine your picture.
Answer+
(a) (−∞,3]
(b) f(x) in increasing on (−∞,2) and decreasing on (2,3). There is a local maximum at x=2 and a local minimum at the endpoint x=3.
(c) f(x) is always concave down and has no inflection points.
(d) (3,0)
(e)
Full solution+
(a) Since we must have 3−x≥0, this tells us x≤3.
So, the domain is (−∞,3].
(b)
f′(x)=3−x−23−xx=323−x2−x
For every x in the domain of f′(x), the denominator is positive, so the sign of f′(x) depends only on the numerator.
x
(−∞,2)
2
(2,3)
3
f′(x)
positive
0
negative
DNE
f(x)
increasing
maximum
decreasing
endpoint
So, f is increasing for x<2, has a local (in fact global) maximum at x=2,
is decreasing for 2<x<3, and has a local minimum at x=3.
Remark: this shows us the basic skeleton of the graph. It consists of a single hump.
(c) When x<3,
f′′(x)=41(3x−12)(3−x)−3/2<0
The domain of f′′(x) is (−∞,3), and over its domain it is always negative
(the factor
(3x−12) is negative for all x<4 and the factor
(3−x)−3/2 is positive for all x<3). So, f(x) has no inflection points and is concave
down everywhere.
(d) We already found
f′(x)=323−x2−x.
This is undefined at x=3. Indeed,
x→3−lim323−x2−x=−∞,
so
f(x) has a vertical tangent line at (3,0).
(e)
To sketch the curve y=f(x), we already know its intervals of increase and decrease, and its concavity. We also note its intercepts are (0,0) and (3,0).
In Questions 2 through 4, you will sketch the graphs of rational functions.
Indicate the critical points, local and absolute maxima and minima,
vertical and horizontal asymptotes, inflection points and regions where the
curve is concave upward or downward.
Hint+
The local maximum is also a global maximum.
Answer+
The square mark is the inflection point, and the closed dot is the local and global maximum.
Full solution+
Asymptotes:
x→±∞limf(x)=x→±∞limx1−x42=0
So y=0 is a horizontal asymptote both at x=∞ and
x=−∞.
x→0limf(x)=x→0limx4x3−2=−∞
So there is a vertical asymptote at x=0, where the function plunges downwards from both the right and the left.
Intervals of increase and decrease:
f′(x)=−x21+x58=x58−x3
The only place where f′(x) is zero only at x=2. So f(x) has a horizontal tangent at x=2,
y=83. This is a critical point.
The derivative is undefined at x=0, as is the function.
x
(−∞,0)
0
(0,2)
2
(2,∞)
f′(x)
negative
DNE
positive
0
negative
f(x)
decreasing
vertical asymptote
increasing
local max
decreasing
Since the function changes from increasing to decreasing at x=2, the only local maximum is at x=2.
At this point, we get a rough sketch of f(x).
Concavity:
f′′(x)=x32−x640=x62x3−40
The second derivative of f(x) is positive for x>320 and negative for x<320.
So the curve is concave up for x>320 and concave down for
x<320. There is an inflection point at x=320≈2.7,
y=204/318≈0.3.
Intercepts:
Since f(x) is not defined at x=0, there is no y-intercept. The only x-intercept is x=32≈1.3.
Sketch:
We can add concavity to our skeleton sketched above, and label our intercept and inflection point (marked with a square).
The first and second derivatives of the function f(x)=1+x3x4 are:
f′(x)=(1+x3)24x3+x6andf′′(x)=(1+x3)312x2−6x5
Graph f(x). Include local
and absolute maxima and minima, regions where f(x) is increasing or
decreasing, regions where the curve is concave upward
or downward, and any asymptotes.
Hint+
The sign of the first derivative is determined entirely by the numerator, but the sign of the second derivative depends on both the numerator and the denominator.
Answer+
The square mark marks the inflection point.
Full solution+
Asymptotes:
When x=−1, the denominator 1+x3 of f(x) is zero while the numerator is 1, so x=−1 is a vertical asymptote. More precisely,
x→−1−limf(x)=−∞x→−1+limf(x)=∞
There are no horizontal asymptotes, because
x→∞lim1+x3x4=∞x→−∞lim1+x3x4=−∞
Intervals of increase and decrease:
We note that f′(x) is defined for all x=−1 and is not defined for x=−1. Therefore, the only singular point for f(x) is x=−1.
At these critical points, f(0)=0 and
f(−34)=−3434<0.
The denominator of f′(x) is never negative, so the sign of f′(x) is the same as the sign of its numerator, x3(4+x3).
x
(−∞,−34)
−34
(−34,−1)
−1
(−1,0)
0
(0,∞)
f′(x)
positive
0
negative
DNE
negative
0
positive
f(x)
increasing
l. max
decreasing
VA
decreasing
l. min
increasing
Now, we have enough information to make a skeleton of our graph.
Concavity:
The second derivative is undefined when x=−1. It is zero when 12x2−6x5=6x2(2−x3)=0.
That is, at x=32≈1.3 and x=0. Notice that the sign of f′′(x) does not change at x=0, so x=0 is not an inflection point.
x
(−∞,−1)
−1
(−1,0)
0
(0,32)
32
(32,∞)
f′′(x)
negative
DNE
positive
0
positive
0
negative
f(x)
concave down
VA
concave up
concave up
IP
concave down
Now we can refine our skeleton by adding concavity.
The only intercept of f(x) is the origin. In particular, that means that out of the three intervals where it is continuous,
namely (−∞,−1), (−1,1) and (1,∞), in two of them f(x) is always positive or always negative.
When x<−1: 1−x2<0 and x3<0, so f(x)>0.
When x>1: 1−x2<0 and x3>0, so f(x)<0.
When −1<x<0, 1−x2>0 and x3<0 so f(x)<0.
When 0<x<1, 1−x2>0 and x3>0 so f(x)>0.
Intervals of increase and decrease:
f′(x)=(1−x2)23x2−x4=(1−x2)2x2(3−x2)
The only singular points are x=±1, where f(x), and hence f′(x), is not defined. The critical points are:
f′(x)x2=0x=0=0or3−x2=0orx=±3≈±1.7
The values of f at its critical points are
f(0)=0, f(3)=−233≈−2.6 and f(−3)=233≈2.6.
Notice the sign of f′(x) is the same as the sign of 3−x2.
x
(−∞,−3)
−3
(−3,−1)
−1
f′(x)
negative
0
positive
DNE
f(x)
decreasing
local min
increasing
VA
x
(−1,0)
0
(0,3)
3
(3,∞)
f′(x)
positive
0
positive
0
negative
f(x)
increasing
increasing
local max
decreasing
Now we have enough information to sketch a skeleton of f(x).
Concavity:
f′′(x)=(1−x2)32x(3+x2)
The second derivative is zero when x=0, and is undefined when x=±1.
Determine all of the following if they are present:
x–coordinates of local maxima and minima, intervals
where f(x) is increasing or decreasing;
intervals where f(x) is concave upwards or downwards;
equations of any horizontal or vertical asymptotes.
Sketch the graph of y=f(x), giving the (x,y)
coordinates for all points of interest above.
Hint+
The function is continuous at x=0, but its derivative is not.
Answer+
(a)
One branch of the function,
the exponential function ex, is continuous everywhere. So f(x) is continuous for x<0. When x≥0, f(x)=3(x+1)x2+3, which is continuous whenever x=−1 (so it's continuous for all x>0). So, f(x) is continuous for x>0. To see that f(x) is continuous at x=0, we see:
Hence f(x) is continuous at x=0, so f(x) is continuous everywhere.
(b)
i. f(x) is increasing for x<0 and x>1, decreasing for 0<x<1,
has a local max at (0,1), and has a local min at (1,32).
ii. f(x) is concave upwards for all x=0.
iii.
The x–axis is a horizontal asymptote as x→−∞.
(c)
Full solution+
(a)
One branch of the function,
the exponential function ex, is continuous everywhere. So f(x) is continuous for x<0. When x≥0, f(x)=3(x+1)x2+3, which is continuous whenever x=−1 (so it's continuous for all x>0). So, f(x) is continuous for x>0. To see that f(x) is continuous at x=0, we see:
i.
The only singular point is x=0, and the only critical point is x=1. (When you're reading off the expression for f′(x), remember that the bottom line only applies when x>0.)
x
(−∞,0)
0
(0,1)
1
(1,∞)
f′(x)
positive
DNE
negative
0
positive
f(x)
increasing
local max
decreasing
local min
increasing
The coordinates of the local maximum are (0,1)
and the coordinates of the local minimum are (1,32).
ii.
When x=0, f′′(x) is always positive, so f(x) is concave up.
iii.
So, there is no horizontal asymptote to the right.
x→−∞limf(x)=x→−∞limex=0
So, y=0 is a horizontal asymptote to the left.
Since f(x) is continuous everywhere, there are no vertical asymptotes.
(c)
In Questions 6 and 7, you will sketch the graphs of functions with an exponential component. In the next section, you will learn how to find their horizontal asymptotes, but for now these are given to you.
The function f(x) and its derivative are given below:
f(x)=(1+2x)e−x2andf′(x)=2(1−x−2x2)e−x2
Sketch the graph of f(x). Indicate the critical points, local
and/or absolute maxima and minima, and asymptotes. Without actually calculating
the inflection points, indicate on the graph their approximate location.
Note: x→±∞limf(x)=0.
Hint+
Since you aren't asked to find the intervals of concavity exactly, sketch the intervals of increase and decrease, and turn them into a smooth curve. You might not get exactly the intervals of concavity that are given in the solution, but there should be the same number of intervals as the solution, and they should have the same positions relative to the local extrema.
Answer+
Full solution+
Asymptotes:
In the problem statement, we are told:
x→±∞limex21+2x=0
So, y=0 is a horizontal asymptote both at x=∞ and at x=−∞.
Since f(x) is continuous, it has no vertical asymptotes.
Intervals of increase and decrease:
The critical points are the zeroes of 1−x−2x2=(1−2x)(1+x). That is,
x=21,−1.
x
(−∞,−1)
−1
(−1,21)
21
(21,∞)
f′(x)
negative
0
positive
0
negative
f(x)
decreasing
local min
increasing
local max
decreasing
At these critical points, f(21)=2e−1/4>0 and f(−1)=−e−1<0.
From here, we can sketch a skeleton of the graph.
Concavity:
We are told that we don't have to actually solve for the inflection points. We just need to know enough to get a basic idea. So, we'll turn the skeleton of the graph into smooth curve.
Inflection points are points where the convexity changes from up
to down or vice versa. It looks like our graph is convex down for x from
−∞ to about −1.8, convex up from about x=−1.8 to about x=−0.1,
convex down from about x=−0.1 to about x=1.4 and convex up from about
x=1.4 to infinity. So there are three inflection points at roughly
x=−1.8,−0.1,1.4.
Find all inflection points and intervals of increase, decrease,
convexity up, and convexity down.
You may use without proof the formula
f′′(x)=(x3−3x)e−x2/2.
Find local and global minima and maxima.
Use all the above to draw a graph for f. Indicate all
special points on the graph.
Hint+
Use intervals of increase and decrease, concavity, and asymptotes to sketch the curve.
(b)
The local and global minimum of f(x) is at (−1,e−1), and the local and global maximum of f(x) is at (1,e1).
(c)
In the graph below, square marks are inflection points, and
solid dots are extrema.
Full solution+
(a) We need to know the first and second derivative of f(x). Using the product and chain rules, f′(x)=e−x2/2(1−x2). Given to us is f′′(x)=(x3−3x)e−x2/2. (These derivatives are also easy to find using the formula developed in Question 20, Section 3.4.)
Since e−x2/2 is always positive, the sign of f′(x) is the same as the sign of 1−x2. f(x) has no singular points and its only critical points are x=±1. At these critical points, f(−1)=−e1 and f(1)=e1.
x
(−∞,−1)
−1
(−1,1)
1
(1,∞)
f′(x)
negative
0
positive
0
negative
f(x)
decreasing
local min
increasing
local max
decreasing
This, together with the observations that f(x)<0 for x<0,
f(0)=0 and f(x)>0 for x>0 (in fact f is an odd function),
is enough to sketch a skeleton of our graph.
We can factor f′′(x)=(x3−3x)e−x2/2=x(x+3)(x−3)e−x2/2. Since e−x2/2 is always positive, the sign of f′′(x) is the same as the sign of x(x+3)(x−3).
x
(−∞,−3)
−3
(−3,0)
0
(0,3)
3
(3,∞)
f′′(x)
negative
0
positive
0
negative
0
positive
f(x)
concave down
IP
concave up
IP
concave down
IP
concave up
(b) We've already seen that f(x) has a local min at x=−1 and a local max at x=1.
As x tends to negative infinity, f(x) tends to 0, and f(x) is decreasing on (−∞,−1). Then f(x) is between 0 and f(−1)=e−1 on (−∞,−1). Then f(x) is increasing on (−1,1) from f(−1)=e−1 to f(1)=e1. Finally, for x>1, f(x) is decreasing from f(1)=e1 and tending to 0. So when x>1, f(x) is between e1 and 0.
So, over its entire domain, f(x) is between e−1 and e1, and it only achieves those values at x=−1 and x=1, respectively. Therefore, the local and global min of f(x) is at (−1,e−1), and the local and global max of f(x) is at (1,e1).
(c)
In the graph below, square marks are inflection points, and
solid dots are extrema.
In Questions 8 and 9, you will sketch the graphs of functions that have a trigonometric component.
Use the techniques from this section to sketch the graph of f(x)=x+2sinx.
Hint+
Although the function exhibits a certain kind of repeating behaviour, it is not periodic.
Answer+
Local maxima occur at x=32π+2πn for all integers n, and local minima occur at x=−32π+2πn for all integers n. Inflection points occur at
every integer multiple of π.
Full solution+
Symmetry:
f(−x)=−x+2sin(−x)=−x−2sinx=−f(x)
So, f(x) is an odd function. If we can sketch y=f(x) for nonnegative x, we can use symmetry to complete the curve for all x.
Asymptotes:
Since f(x) is continuous, it has no vertical asymptotes. It also has no horizontal asymptotes, since
x→−∞limf(x)=−∞x→∞limf(x)=∞
Intervals of increase and decrease:
Since f(x) is differentiable everywhere, there are no singular points.
f′(x)=1+2cosx
So, the critical points of f(x) occur when
cosxx=−21=2πn±32π for any integer n
For instance, f(x) has critical points at x=32π, x=34π, x=38π, and x=310π.
From the unit circle, or the graph of y=1+2cosx, we see:
x
(−32π,32π)
32π
(32π,34π)
34π
(34π,38π)
38π
(38π,310π)
f′(x)
positive
0
negative
0
positive
0
negative
f(x)
increasing
l. max
decreasing
l. min
increasing
l. max
decreasing
We have enough information to sketch a skeleton of the curve y=f(x). We use the pattern above for the graph to the right of the y-axis, and use odd symmetry for the graph to the left of the y-axis.
Concavity:
f′′(x)=−2sinx
So, f′′(x) exists everywhere, and is zero for x=π+πn for every integer n.
x
(0,π)
π
(π,2π)
2π
(2π,3π)
3π
(3π,4π)
f′′(x)
negative
0
positive
0
negative
0
positive
f(x)
concave down
IP
concave up
IP
concave down
IP
concave up
Using these values, and the odd symmetry of f(x), we can refine our skeleton. The closed dots are local extrema, and the square marks are inflection points occurring at every integer multiple of π.
Graph the equation y=f(x),
including all important features. (In particular, find all local maxima and
minima and all inflection points.)
Additionally, find the maximum and minimum values of f(x) on the interval [0,π].
Hint+
The period of this function is 2π. So, it's enough to graph the curve y=f(x) over the interval [−π,π], because that figure will simply repeat.
Use trigonometric identities to write f′′(x)=−4(4sin2x+sinx−2). Then you can find where f′′(x)=0 by setting y=sinx and solving 0=4y2+y−2.
Answer+
Below is the graph y=f(x) over the interval [−π,π]. The sketch of the curve over a larger domain is simply a repetition of this figure.
On the interval [0,π], the maximum value of f(x) is 6 and the minimum value is −2.
Let a=sin−1(8−1+33)≈0.635≈0.2π and
b=sin−1(8−1−33)≈−1.003≈−0.3π. The points −π−b, b, a, and π−a are inflection points.
Full solution+
We first compute the derivatives f′(x) and f′′(x).
However, these are not the only values of x in [−π,π] with sinx=8−1±33.
The analysis above misses the others because the arcsine
function only returns numbers in the range [−2π,2π]. The graph below shows that there should be other values of x with sinx=8−1±33, and hence f′′(x)=0.
We can recover the other solutions in [−π,π] by recalling that
sin(x)=sin(π−x).
So, if we choose x=arcsin(8−1+33)≈0.635 to make
sin(x)=8−1+33 so that f′′(x)=0, then setting
x=π−a=π−arcsin(8−1+33)≈2.507
will also give us sin(x)=8−1+33 and f′′(x)=0. Similarly, setting
x=π−b=π−arcsin(8−1−33)≈4.145
would give us f′′(x)=0. However, this value is outside [−π,π]. To find another solution inside [−π,π] we
use the identity
sin(x)=sin(−π−x)
(which we can obtain from the identity we used above and the fact that sin(θ)=sin(θ±2π) for any angle θ). Using this,
we can show that
x=−π−b=−π−arcsin(8−1−33)≈−2.139
also gives f′′(x)=0.
So, all together, f′′(x)=0 when x=−π−b, x=b, x=a, and x=π−a.
Now, we should compute the sign of f′′(x) while x is between −π and π. Recall that, if y=sinx, then f′′(x)=−4(4y2+y−2). So, in terms of y, f′′ is a parabola pointing down, with intercepts y=8−1±33. Then f′′ is positive when y is in the interval (8−1−33,8−1+33), and f′′ is negative otherwise. From the graph of sine, we see that y is between
8−1−33 and 8−1+33 precisely on the intervals (−π,−π−b),
(b,a), and (π−a,π).
Therefore,
f(x) is concave up on the intervals (−π,−π−b), (b,a), and (π−a,π), and
f(x) is concave down on the intervals (−π−b,b) and (a,π−a). So, the inflection points of f occur at
x=−π−b, x=b, x=a, and x=π−a.
To find the maximum and minimum values of f(x) on [0,π], we compare the values of f(x) at its critical points in this interval (only x=2π) with the values of f(x) at its endpoints x=0, x=π.
Since f(0)=f(π)=−2, the minimum value
of f on [0,π] is −2, achieved at x=0,π and the maximum
value of f on [0,π] is 6, achieved at x=2π.
You may use the facts y′(x)=3x5/3(x+1)2/3−(x+2) and
y′′(x)=9x8/3(x+1)5/34x2+16x+10.
Hint+
There is one point where the curve is continuous but has a vertical tangent line.
Answer+
The closed dot is the local minimum, and the square marks are inflection points at x=−1 and x=−2±1.5. The graph has horizontal asymptotes y=0 as x goes to ±∞.
Full solution+
Let f(x)=3x2x+1.
Asymptotes: Since x→0limf(x)=∞, f(x) has a vertical asymptote at x=0 where the curve reaches steeply upward from both the left and the right.
x→±∞limf(x)=0, so y=0 is a horizontal asymptote for x→±∞.
Intercepts: f(−1)=0.
Intervals of increase and decrease:
f′(x)=3x5/3(x+1)2/3−(x+2)
There is a singular point at x=−1 and a critical point at x=−2, in addition to a discontinuity at x=0. Note that (x+1)2/3=(3x+1)2, which is never negative. Note also that x→−1limf′(x)=∞, so f(x) has a vertical tangent line at x=−1.
x
(−∞,−2)
−2
(−2,−1)
−1
(−1,0)
0
(0,∞)
f′(x)
negative
0
positive
DNE
positive
DNE
negative
f(x)
decreasing
l. min
increasing
vertical
increasing
VA
decreasing
This gives us enough information to sketch a skeleton of the curve.
Concavity:
f′′(x)=9x8/3(x+1)5/34x2+16x+10
We still have a discontinuity at x=0, and f′′(x) does not exist at x=−1. The second derivative is zero when 4x2+16x+10=0. Using the quadratic formula, we find this occurs when x=−2±1.5≈−0.8,−3.2. Note x8/3=(3x)8 is never negative.
x
(−∞,−2−1.5)
−2−1.5
(−2−1.5,−1)
−1
f′′(x)
negative
0
positive
DNE
f(x)
concave down
IP
concave up
IP
x
(−1,−2+1.5)
−2+1.5
(−2+1.5,0)
(0,∞)
f′′(x)
negative
0
positive
positive
f(x)
concave down
IP
concave up
concave up
Now, we can refine our skeleton. The closed dot is the local minimum, and the square marks are inflection points.
A function f(x) defined on the whole real number line satisfies
the following conditions
f(0)=0f(2)=2x→+∞limf(x)=0f′(x)=K(2x−x2)e−x
for some positive constant K. (Read carefully: you are given the derivative of f(x), not f(x) itself.)
Determine the intervals on which f is increasing and decreasing
and the location of any local maximum and minimum values of f.
Determine the intervals on which f is concave up or down
and the x–coordinates of any inflection points of f.
Determine x→−∞limf(x).
Sketch the graph of y=f(x), showing any asymptotes and the
information determined in parts (a) and (b).
Hint+
Use x→−∞limf′(x) to determine x→−∞limf(x).
Answer+
(a) decreasing for x<0 and x>2, increasing for 0<x<2, minimum at (0,0), maximum at (2,2).
(b) concave up for x<2−2 and x>2+2, concave down for 2−2<x<2+2, inflection points at x=2±2.
(c)∞
(d)
Square marks indicate inflection points, and closed dots indicate local extrema.
Full solution+
The parts of the question are just scaffolding to lead you through sketching the curve. Their answers are given explicitly, in an organized manner, in the “answers" section. In this solution, they are scattered throughout.
Asymptotes:
Since the function has a derivative at every real number, the function is continuous for every real number, so it has no vertical asymptotes. In the problem statement, you are told x→∞limf(x)=0, so y=0 is a horizontal asymptote as x goes to infinity. It remains to evaluate x→−∞limf(x). Let's consider the limit of f′(x) instead. Recall K is a positive constant.
x→−∞lime−x=x→∞limex=∞x→−∞limK(2x−x2)=−∞
So,
x→−∞limK(2x−x2)e−x=−∞
That is, as x becomes a hugely negative number, f′(x) also becomes a hugely negative number. As we move left along the x-axis, f(x) is decreasing with a steeper and steeper slope, as in the sketch below. That means
x→−∞limf(x)=∞.
Intervals of increase and decrease:
We are given f′(x) (although we don't know f(x)):
f′(x)=Kx(2−x)e−x
The critical points of f(x) are x=0 and x=2, and there are no singular points. Recall e−x is always positive, and K is a positive constant.
x
(−∞,0)
0
(0,2)
2
(2,∞)
f′(x)
negative
0
positive
0
negative
f(x)
decreasing
local min
increasing
local max
decreasing
So, f(0)=0 is a local minimum, and f(2)=2 is a local maximum.
Looking ahead to part (d), we have a skeleton of the curve.
Sketch the graph of the equation y=f(x). Indicate any
local extrema and inflection points.
Sketch the graph of the inverse function y=g(x)=f−1(x).
Find the domain and range of the inverse function g(x)=f−1(x).
Evaluate g′(21).
Hint+
Once you have the graph of a function, reflect it over the line y=x to graph its inverse.
Be careful of the fact that f(x) is only defined in this problem for x≥0.
Answer+
(a)
There are no inflection points or extrema, except the endpoint (0,1).
(b)
There are no inflection points or extrema, except the endpoint (1,0).
(c)
The domain of g is (0,1].
The range of g is [0,∞).
(d)
g′(21)=−2
Full solution+
(a)
You should be familiar with the graph of y=ex. You can
construct the graph of y=e−x just by reflecting the graph
of y=ex across the y–axis. To see why this is the case,
imagine swapping each value of x with its negative: for example,
swapping the point at x=−1 with the point at x=1, etc.
Alternatively, you can graph y=f(x)=e−x, x≥0,
using the methods of this section: at x=0, y=f(0)=1;
as x increases, y=f(x)=e−x decreases, with no local
extrema; and as x→+∞, y=f(x)→0.
There are no inflection points or extrema, except the endpoint (0,1).
(b)
Recall that, to graph the inverse of a function, we reflect the original function across the
line y=x. To see why this is true, consider the following. By definition, the inverse function
g of f is obtained by solving y=f(x) for x as a function of y. So, for any pair of
numbers x and y, we have
f(x)=y if and only g(y)=x
That is, g is the function that swaps the input and output of f. Now the point (x,y)
lies on the graph of f if and only if y=f(x). Similarly, the point (X,Y) lies
on the graph of g if and only if Y=g(X). Choosing Y=x and X=y, we see that the point
(X,Y)=(y,x) lies on the graph of g if and only if x=g(y), which in turn is the case
if and only if y=f(x). So
(y,x) is on the graph of g if and only if (x,y) is on the graph of f.
To get from the point (x,y) to the point (y,x) we have to
exchange x↔y, which we can do by reflecting
over the line y=x. Thus we can construct the graph of g
by reflecting the curve y=f(x) over the line y=x.
(c)
The domain of g is the range of f, which is (0,1].
The range of g is the domain of f, which is [0,∞).
Sketch the graph of y=f(x)=x5−x, indicating asymptotes,
local maxima and minima, inflection points, and where the
graph is concave up/concave down.
Consider the function f(x)=x5−x+k, where k is a constant,
−∞<k<∞. How many roots does the function have? (Your answer might
depend on the value of k.)
Answer+
(a)
Local maximum at x=−451;
local minimum at x=451;
inflection point at the origin;
concave down for x<0 ; concave up for x>0.
(b) The number of distinct real roots of x5−x+k is:
1 when ∣k∣>5454
2 when ∣k∣=5454
3 when ∣k∣<5454
Full solution+
(a)
First, we differentiate.
f(x)=x5−xf′(x)=5x4−1f′′(x)=20x3
The function and its first derivative tells us the following:
x→∞limf(x)=∞, x→−∞limf(x)=−∞
f′(x)>0 (i.e. f is increasing) for
∣x∣>451
f′(x)=0 (i.e. f has critical points) for x=±451≈±0.67
f′(x)<0 (i.e. f is decreasing) for ∣x∣<451
f(±451)=∓5454≈∓0.53
This gives us a first idea of the shape of the graph.
We refine this skeleton using information from the second derivative.
f′′(x)>0 (i.e. f is concave up) for x>0,
f′′(x)=0 (i.e. f has an inflection point) for x=0, and
f′′(x)<0 (i.e. f is concave down) for x<0
Thus
f has no asymptotes
f has a local maximum at x=−451
and a local minimum at x=451
f has an inflection point at x=0
f is concave down for x<0 and concave up for x>0
(b) The function
x5−x+k has a root at x=x0 if and only if x5−x=−k
at x=x0. So the number of distinct real roots of x5−x+k is the
number of times the curve y=x5−x crosses the horizontal line y=−k.
The local maximum of x5−x (when x=−451)
is 5454, and the local minimum of x5−x
(when x=451) is −5454.
So, looking at the graph of x5−x above, we see that the number of distinct
real roots of x5−x+k is
You may use these properties in your solution to this question.
Sketch the graphs of sinh(x) and cosh(x).
Define inverse hyperbolic trigonometric functions
sinh−1(x) and cosh−1(x), carefully specifing their domains
of definition. Sketch the graphs of sinh−1(x) and cosh−1(x).
Find
dxd{cosh−1(x)}.
Hint+
For (a), don't be intimidated by the new names: we can graph these functions using the methods learned in this section.
For (b), remember that to define an inverse of a function, we need to restrict the domain of that function to an interval where it is one-to-one. Then to graph the inverse, we can simply reflect the original function over the line y=x.
For (c), set y(x)=cosh−1(x), so cosh(y(x))=x. The differentiate using the chain rule. To get your final answer in terms of x (instead of y), use the identity cosh2(y)−sinh2(y)=1.
Answer+
(a)
(b)
For any real x, define sinh−1(x) to be
the unique solution of sinh(y)=x. For every x∈[1,∞), define cosh−1(x) to be the
unique y∈[0,∞) that obeys cosh(y)=x.
(c) dxd{cosh−1(x)}=x2−11
Full solution+
(a)
You might not be familiar with hyperbolic sine and cosine, but you don't need to be. We can graph them using the same methods as the other curves in this section. The derivatives are given to us:
sinh(x) has a derivative that is always positive, so sinh(x) is always increasing. The second derivative of sinh(x) is negative to the left of x=0 and positive to the right of x=0, so sinh(x) is concave down to the left of the y-axis and concave up to its right, with an inflection point at x=0.
cosh(x) has a derivative that is positive when x>0 and negative when x<0. The second derivative of cosh(x) is always positive, so it is always concave up.
cosh(0)=1 and sinh(0)=0.
$\ds\lim_{x\rightarrow\infty}\sinh x = \ds\lim_{x\rightarrow\infty}\cosh x
=\ds\lim_{x\rightarrow\infty}\frac{e^x}{2}=\infty,since\ds\lim_{x \to \infty}e^{-x}=0$
$\ds\lim_{x \to -\infty}\sinh x =\ds\lim_{x \to -\infty}\left(\frac{e^x}{2}-\frac{e^{-x}}{2}\right)
=\ds\lim_{x \to \infty}\left(\frac{e^{-x}}{2}-\frac{e^x}{2}\right)=-\infty$ and
$\ds\lim_{x \to -\infty}\cosh x =\ds\lim_{x \to -\infty}\left(\frac{e^x}{2}+\frac{e^{-x}}{2}\right)
=\ds\lim_{x \to \infty}\left(\frac{e^{-x}}{2}+\frac{e^x}{2}\right)=\infty$
cosh(x) is even, since cosh(−x)=2e−x+e−(−x)=2e−x+ex=cosh(x), and sinh(x) is odd, since sinh(−x)=2e−x−e−(−x)=2e−x−ex=2−(ex−e−x)=−sinh(x)
(b)
As y runs over (−∞,∞) the function
sinh(y) takes every real value exactly once. So,
for each x∈(−∞,∞), define sinh−1(x) to be
the unique solution of sinh(y)=x.
As y runs over [0,∞) the function
cosh(y) takes every real value in [1,∞) exactly once.
In particular, the smallest value of cosh(y) is cosh(0)=1.
So, for each x∈[1,∞), define cosh−1(x) to be the
unique y∈[0,∞) that obeys cosh(y)=x.
To graph the inverse of a (one-to-one) function, we reflect the original function over the line y=x. Using this method to graph y=sinh−1(x) is straightforward. To graph y=cosh(x), we need to be careful of the domains: we are restricting cosh(x) to values of x in [0,∞).
The graphs are
(c) Let y(x)=cosh−1(x). Then, using the definition of cosh−1,
coshy(x)=x
We differentiate with respect to x using the chain rule.
dxd{coshy(x)}y′(x)sinhy(x)=dxd{x}=1
We solve for y′(x).
y′(x)=sinhy(x)1
We want to have our answer in terms of x, not y. We know that coshy=x, so if we can convert hyperbolic sine into hyperbolic cosine, we can get rid of y. Our tool for this is the identity, given in the question statement, cosh2(x)−sinh2(x)=1. This tells us sinh2(y)=1−cosh2(y).
Now we have to decide whether sinh(y) is the positive or negative square root of 1−cosh2(y) in our context. Looking at the graph of y(x)=cosh−1(x), we see y′(x)>0. So we use the positive square root:
y′(x)=cosh2y(x)−11=x2−11
Remark: dxd{arccos(x)}=1−x2−1, so again the hyperbolic trigonometric function has properties similar to (but not exactly the same as) its trigonometric counterpart.
