### Subsection3.4.2First Approximation — the Linear Approximation

Our first  4 Recall that we started counting from zero. approximation improves on our zeroth approximation by allowing the approximating function to be a linear function of $x$ rather than just a constant function. That is, we allow $F(x)$ to be of the form $A+Bx\text{,}$ for some constants $A$ and $B\text{.}$

To ensure that $F(x)$ is a good approximation for $x$ close to $a\text{,}$ we still require that $f(x)$ and $F(x)$ have the same value at $x=a$ (that was our zeroth approximation). Our additional requirement is that their tangent lines at $x=a$ have the same slope — that the derivatives of $f(x)$ and $F(x)$ are the same at $x=a\text{.}$ Hence

\begin{align*} F(x)&=A+Bx & &\implies & F(a)=A+Ba&=f(a)\\ F'(x)&=B & &\implies & F'(a)=\phantom{A+a}B&=f'(a) \end{align*}

So we must have $B=f'(a)\text{.}$ Substituting this into $A+Ba=f(a)$ we get $A=f(a)-af'(a)\text{.}$ So we can write

\begin{align*} F(x) &= A+Bx = \overbrace{f(a)- af'(a)}^A+ f'(a) \cdot x\\ &= f(a) + f'(a) \cdot(x-a) \end{align*}

We write it in this form because we can now clearly see that our first approximation is just an extension of our zeroth approximation. This first approximation is also often called the linear approximation of $f(x)$ about $x=a\text{.}$

We should again stress that in order to form this approximation we need to know $f(a)$ and $f'(a)$ — if we cannot compute them easily then we are not going to be able to proceed.

Recall, from Theorem 2.3.4, that $y=f(a)+f'(a)(x-a)$ is exactly the equation of the tangent line to the curve $y=f(x)$ at $a\text{.}$ Here is a figure showing the graphs of a typical $f(x)$ and the approximating function $F(x)\text{.}$

Observe that the graph of $f(a)+f'(a)(x-a)$ remains close to the graph of $f(x)$ for a much larger range of $x$ than did the graph of our constant approximation, $f(a)\text{.}$ One can also see that we can improve this approximation if we can use a function that curves down rather than being perfectly straight. That is our next approximation.

But before then, back to our example:

Use the linear approximation to estimate $e^{0.1}\text{.}$

Solution First set $f(x) = e^x$ and $a=0$ as before.

• To form the linear approximation we need $f(a)$ and $f'(a)\text{:}$
\begin{align*} f(x) &= e^x & f(0) & = 1\\ f'(x) &= e^x & f'(0) & = 1 \end{align*}
• Then our linear approximation is
\begin{align*} F(x) &= f(0) + x f'(0) = 1 + x\\ F(0.1) &= 1.1 \end{align*}

Recall that $e^{0.1} = 1.105170918\dots\text{,}$ so the linear approximation is almost correct to 3 digits.

It is worth doing another simple example here.

Use a linear approximation to estimate $\sqrt{4.1}\text{.}$

Solution First set $f(x)=\sqrt{x}\text{.}$ Hence $f'(x) = \frac{1}{2\sqrt{x}}\text{.}$ Then we are trying to approximate $f(4.1)\text{.}$ Now we need to choose a sensible $a$ value.

• We need to choose $a$ so that $f(a)$ and $f'(a)$ are easy to compute.

• We could try $a=4.1$ — but then we need to compute $f(4.1)$ and $f'(4.1)$ — which is our original problem and more!
• We could try $a=0$ — then $f(0)=0$ and $f'(0) = DNE\text{.}$
• Setting $a=1$ gives us $f(1)=1$ and $f'(1)=\frac{1}{2}\text{.}$ This would work, but we can get a better approximation by choosing $a$ is closer to $4.1\text{.}$
• Indeed we can set $a$ to be the square of any rational number and we'll get a result that is easy to compute.
• Setting $a=4$ gives $f(4)=2$ and $f'(4) = \frac{1}{4}\text{.}$ This seems good enough.
• Substitute this into equation 3.4.3 to get
\begin{align*} f(4.1) &\approx f(4) + f'(4) \cdot(4.1-4)\\ &= 2 + \frac{0.1}{4} = 2 + 0.025 = 2.025 \end{align*}

Notice that the true value is $\sqrt{4.1} = 2.024845673\dots\text{.}$