Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 8

Check the Linear Approximation $\tan x \approx x$ at $a = 0$. Then determine the values of $x$ for which the Linear Approximation is accurate to within $0.1$.

Let $f(x) = \tan x$

Using the Linear Approximation/Tangent Line Approximation

$L(x) = f(a) + f'(a)(x - a)$


$
\begin{equation}
\begin{aligned}

f(a) = f(0) =& \tan (0)
\\
\\
f(0) =& 0
\\
\\
f'(a) = f'(0) =& \frac{d}{dx} (\tan x)
\\
\\
f'(0) =& \sec ^2 x
\\
\\
f'(0) =& \sec ^2 (0)
\\
\\
f'(0) =& \frac{1}{\cos ^2 (0)}
\\
\\
f'(0) =& 1
\\
\\
L(x) =& 0 + 1 (x - 0)
\\
\\
L(x) =& x

\end{aligned}
\end{equation}
$


So

$\tan x \approx x$

Accuracy to within $0.1$ means that the function should differ by less than $0.1$

$|\tan x - x| < 0.1$

Equivalently, we have

$\tan x - 0.1 < x < \tan x + 0.1$







This says that the Linear Approximation should lie between the curves obtained by shifting the curve $y = \tan x$ upward and downward by $0.1$. The graph shows the tangent line $y = x$ intersects both the upper and lower curve at A and B. We can estimate the $x$-coordinate of A which is $-0.63$ and the $x$-coordinate of B is $0.63$.

Thus, referring to the graph the approximation

$\tan x \approx x$

is accurate to within $0.1$ when $-0.63 < x < 0.63$