8.1 Local and global maxima and minima

Estimate $f'(0)$.

There is a critical point at $x=0$. The $x$-value of the red dot is a singular point, and a local maximum occurs there.

When $x=0$, the curve $y=f(x)$ appears to have a flat tangent line, so the $x=0$ is a critical point. However, it is not a local extremum: it is not true that $f(0) \geq f(x)$ for all $x$ near 0, and it is not true that $f(0) \leq f(x)$ for all $x$ near 0.

To the right of the $x$-axis, there is a spike where the derivative of $f(x)$ does not exist. The $x$-value corresponding to this spike (call it $a$) is a singular point, and $f(x)$ has a local maximum at $x=a$.

If the graph is discontinuous at a point, it is not differentiable at that point.

The $x$-coordinate corresponding to the blue dot (let's call it $a$) is a critical point, and $f(x)$ has a local and global minimum at $x=a$. The $x$-coordinate corresponding to the discontinuity (let's call it $b$) is a singular point, but there is not a global or local extremum at $x=b$.

The $x$-coordinate corresponding to the blue dot (let's call it $a$) is a critical point, because the tangent line to $f(x)$ at $x=a$ is horizontal. There is no lower point nearby, and actually no lower point on the whole interval shown, so $f(x)$ has both a local minimum and a global minimum at $x=a$.

If a function is not continuous at a point, then it is not differentiable at that point. So, the $x$-coordinate corresponding to the discontinuity (let's call it $b$) is a singular point. Values of $f(x)$ immediately to the right of $b$ are lower, and values immediately to the left of $b$ are higher, so $f(x)$ has no local (or global) extremum at $x=b$.

Try making a little bump at $x=2$, the letting the function get quite large somewhere else.

One possible answer is shown below.

One possible answer is shown below.

For every $x$ in the red interval shown below, $f(2) \geq f(x)$, so $f(2)$ is a local maximum. However, the point marked with a blue dot shows that $f(x)>f(2)$ for some $x$, so $f(2)$ is not a global maximum.

Critical points are those values of $x$ for which $f'(x)=0$.

Singular points are those values of $x$ for which $f(x)$ is not differentiable.

The critical points are $x=3$ and $x=-1$. These two points are the only places where local extrema might exist. There are no singular points.

Critical points are those values of $x$ for which $f'(x)=0$, and singular points are those values of $x$ for which $f(x)$ is not differentiable. So, we ought to find $f'(x)$. Using the quotient rule,

(a) The derivative $f'(x)$ is zero when $x=3$ and when $x=-1$, so those are the critical points.

(b) The denominator of $f'(x)$ is never zero, so the derivative $f'(x)$ exists for all $x$ and $f(x)$ has no singular points.

(c) Theorem 8.1.3 tells us that local extrema of $f(x)$ can only occur at critical points and singular points. So, the possible points where extrema of $f(x)$ may exist are $x=3$ and $x=-1$.

We're only after local extrema, not global. Let $f(x)$ be our function. If there is some interval around $x=2$ where nothing is bigger than $f(2)$, then $f(2)$ is a local maximum, whether or not it is a maximum overall.

For the first curve, the function's value at $x=2$ (that is, the $y$-value of the solid dot) is higher than anything around it. So, it's a local maximum.

For the second curve, the function's value at $x=2$ (that is, the $y$-value of the solid dot) is higher than everything to the left, but lower than values immediately to the right. (On the graph reproduced below, $f(x)$ is higher than everything in the red section, and lower than everything in the blue section.) So, it is neither a local max nor a local min.

Similarly, for the third curve, $f(2)$ is lower than the values to the right of it, and higher than values to the left of it, so it is neither a local minimum nor a local maximum.

In the final curve, $f(2)$ (remember–this is the $y$-value of the solid dot) is higher than everything immediately to the left or right of it (for instance, over the interval marked in red below), so it is a local maximum.

By Theorem 8.1.3, if $x=2$ not a critical point, then it must be a singular point.

There are many possible answers. Every answer must have $x=2$ as a singular point strictly inside the domain of $f(x)$. Two possibilities are shown below.

The question specifies that $x=2$ must not be an endpoint. By Theorem 8.1.3, if $x=2$ not a critical point, then it must be a singular point. That is, $f(x)$ is not differentiable at $x=2$. Two possibilities are shown below, but there are infinitely many possible answers.

You should be able to figure out the global minima of $f(x)$ in your head.

Remember with absolute values, $|X|=\left{\begin{array}{ll} X&X\ge0\ -X&X<0 \end{array}\right.$.

$x=-7$, $x=-1$, and $x=5$

Critical points are those values of $x$ for which $f'(x)=0$, and singular points are those values of $x$ for which $f(x)$ is not differentiable. So, we ought to find $f'(x)$. Since $f(x)$ has an absolute value sign, let's re-write it in a version that is friendlier to differentiation. Remember that $|X|=X$ when $X \geq 0$, and $|X|=-X$ when $X<0$.

The product $(x-5)(x+7)$ is positive when $(x-5)$ and $(x+7)$ have the same sign, and negative when they have opposite signs, so

Now, when $x \neq -7,5$, we can differentiate, using the chain rule.

We are tempted to say that the derivative doesn't exist when $x=-7$ and $x=5$, but be careful– we don't actually know that yet. The formulas we have for the $f'(x)$ are only good when $x$ is not $-7$ or $5$.

The middle formula $\dfrac{-2x-2}{2\sqrt{-(x-5)(x+7)}}$ tells us $x=-1$ is a critical point: when $x=-1$, $f'(x)$ is given by the middle line, and it is 0. Note that $x=-1$ also makes the top formula 0, but $f'(-1)$ is not given by the top formula, so that doesn't matter.

What we've concluded so far is that $x=-1$ is a critical point of $f(x)$, and $f(x)$ has no other critical points or singular points when $x \neq -7,5$. It remains to figure out what's going on at $-7$ and $5$. One way to do this is to use the definition of the derivative to figure out what $f'(-7)$ and $f'(5)$ are, if they exist. This is somewhat laborious. Let's look for a better way.

So, all together:

$x=-1$ is a critical point, and $x=-7$ and $x=5$ are critical points or singular points (but we don't know which).

Remark: if you would like a review of how to use the definition of the derivative, below we show that $f(x)$ is not differentiable at $x=-7$. (In fact, $x=-7$ and $x=5$ are both singular points.)

Let's first consider the case $h>0$.

Since one side of the limit doesn't exist,

so $f'(x)$ is not differentiable at $x=-7$. Therefore, $x=-7$ is a singular point.

Review the definitions of critical points and extrema: Definition 7.2.1 and Definition 8.1.2.

Every real number $c$ is a critical point of $f(x)$, and $f(x)$ has a local and global maximum and minimum at $x=c$. There are no singular points.

For any real number $c$, $c$ is in the domain of $f(x)$ and $f'(c)$ exists and is equal to zero. So, following Definition 7.2.1, every real number is a critical point of $f(x)$, and $f(x)$ has no singular points.

For every number $c$, let $a=c-1$ and $b=c+1$, so $a<c<b$. Then $f(x)$ is defined for every $x$ in the interval $[a,b]$, and $f(x) =f(c)$ for every $a \le x \le b$. That means $f(x) \leq f(c)$ and $f(x) \geq f(c)$. So, comparing with Definition 8.1.2, we see that $f(x)$ has a global and local maximum AND minimum at every real number $x=c$.