9.2 Linear approximation

The linear approximation $L(x)$ is chosen so that $f(5)=L(5)$ and $f'(5)=L'(5)$.

(a) $f(5)=6$ (b) $f'(5)=3$ (c) not enough information to know

The linear approximation is $L(x)=3x-9$. Since we're approximating at $x=5$, $f(5)=L(5)$, and $f'(5)=L'(5)$. However, there is no guarantee that $f(x)$ and $L(x)$ have the same value when $x \neq 5$. So:
(a) $f(5)=L(5)=6$
(b) $f'(5)=L'(5)=3$
(c) there is not enough information to find $f(0)$.

The graph of the linear approximation is a line, passing through $(2,f(2))$, with slope $f'(2)$.

The linear approximation is shown in red.

The linear approximation is a line, passing through $(2,f(2))$, with slope $f'(2)$. That is, the linear approximation to $f(x)$ about $x=2$ is the tangent line to $f(x)$ at $x=2$. It is shown below in red.

It's an extremely accurate approximation.

$f(x)=2x+5$

For any constant $a$, $f(a)=(2a+5)$, and $f'(a)=2$, so our approximation gives us

Since $f(x)$ itself is a linear function, the linear approximation is actually just $f(x)$ itself. As a consequence, the linear approximation is perfectly accurate for all values of $x$.

You'll need to centre your approximation about some $x=a$, which should have two properties: you can easily compute $\log(a)$, and $a$ is close to $0.93$.

$\log(0.93) \approx -0.07$

We have no idea what $f(0.93)$ is, but 0.93 is pretty close to 1, and we definitely know $f(1)$. The linear approximation of $f(x)$ about $x=1$ is given by

So, we calculate:

Therefore,

When $x=0.93$:

Approximate the function $f(x) = \sqrt{x}$.

$\sqrt{5} \approx \dfrac{9}{4}$

We approximate the function $f(x)=\sqrt{x}$ about $x=4$, since 4 is a perfect square and it is close to 5.

We estimate $\sqrt{5}\approx\dfrac{9}{4}$.

Remark: $\left(\dfrac{9}{4}\right)^2=\dfrac{81}{16}$, which is pretty close to $\dfrac{80}{16}=5$. Our approximation seems pretty good.

Approximate the function $f(x)=\sqrt[5]{x}$.

$\sqrt[5]{30}\approx \dfrac{79}{40}$

We approximate the function $f(x)=\sqrt[5]{x}$. We need to centre the approximation about some value $x=a$ such that we know $f(a)$ and $f'(a)$, and $a$ is not too far from $30$.

$a$ needs to be a number whose fifth root we know. Since $\sqrt[5]{32}=2$, and 32 is reasonably close to 30, $a=32$ is a great choice.

The linear approximation of $f(x)$ about $x=32$ is

When $x=30$:

We estimate $\sqrt[5]{30}\approx \dfrac{79}{40}$.

Remark: $\dfrac{79}{40} = 1.975$, while $\sqrt[5]{30} \approx 1.97435$. This is a decent estimation.

Approximate the function $f(x)=x^3$.

$10.1^3 \approx 1030$, $10.1^3 = 1030.301$

If $f(x)=x^3$, then $f(10.1)=10.1^3$, which is the value we want to estimate. Let's take the linear approximation of $f(x)$ about $x=10$:

We estimate $10.1^3 \approx 1030$. If we calculate $10.1^3$ exactly (which is certainly possible to do by hand), we get 1030.301.

Remark: in the previous subsection, we used a constant approximation to estimate $10.1^3 \approx 1000$. That approximation was easy to do in your head, in a matter of seconds. The linear approximation is more accurate, but not much faster than simply calculating $10.1^3$.

One possible choice of $f(x)$ is $f(x)=\sin x$.

There are many possible answers. One is $f(x)=\sin x$, $a=0$, and $b=\pi$.

There are many possible answers. One is:

We know that $f(\pi)=0$ and $f(0)=0$. Using a constant approximation of $f(x)$ about $x=0$, our estimation is $f(\pi)\approx f(0)=0$, which is exactly the correct value. However, is we make a linear approximation of $f(x)$ about $x=0$, we get

which is not exactly the correct value.

Remark: in reality, we wouldn't estimate $\sin(\pi)$, because we know its value exactly. The purpose of this problem is to demonstrate that fancier approximations are not always more accurate. At the of this section, we'll talk about how to bound the error of your estimations, to make sure that you are finding something sufficiently accurate.

Compare the derivatives.

$a=\sqrt{3}$

The linear approximation $L(x)$ of $f(x)$ about $x=a$ is chosen so that $L(a)=f(a)$ and $L'(a)=f'(a)$. So,

We've narrowed down $a$ to $\sqrt{3}$ or $-\sqrt{3}$. Recall the linear approximation of $f(x)$ about $x=a$ is $f(a)+f'(a)(x-a),$ so its constant term is $f(a)-af'(a)=\arctan(a)-\dfrac{a}{1+a^2}$. We compute this for $a=\sqrt{3}$ and $a=-\sqrt{3}$.

So, when $a=\sqrt{3}$,

and this does not hold when $a=-\sqrt{3}$. We conclude $a=\sqrt{3}$.