Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 10

Check the Linear Approximation $\displaystyle \frac{1}{\sqrt{4 - x}} \approx \frac{1}{2} + \frac{1}{16} x$ at $****$. Then determine the values of $x$ for which the Linear Approximation is accurate to within $0.1$.

Let $\displaystyle f(x) = \frac{1}{(1 + 2x)^4} x$

Using the Linear Approximation/Tangent Line Approximation

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


$
\begin{equation}
\begin{aligned}

f(a) = f(0) =& \frac{1}{\sqrt{4 - 0}}
\\
\\
f(0) =& \frac{1}{\sqrt{4}}
\\
\\
f(0) =& \frac{1}{2}
\\
\\
f'(a) = f'(0) =& \frac{d}{dx} \left[ \frac{1 }{(4 - x)^{\frac{1}{2}}} \right]
\\
\\
f'(0) =& \frac{\displaystyle (4 - x)^{\frac{1}{2}} \frac{d}{dx} (1) - (1) \frac{d}{dx} (4 - x)^{\frac{1}{2}} }{[(4 x)^{\frac{1}{2}}]^2}
\\
\\
f'(0) =& \frac{\displaystyle (4 - x)^{\frac{1}{2}} (0) - \frac{1}{2} (4 - x)^{\frac{-1}{2}} \frac{d}{dx} (4 - x) }{4 - x}
\\
\\
f'(0) =& \frac{\displaystyle \frac{1}{2} (4 - x)^{\frac{-1}{2}}}{4 -x}
\\
\\
f'(0) =& \frac{1}{2(4 - x)^{\frac{1}{2}} (4 - x)}
\\
\\
f'(0) =& \frac{1}{2(4 - x)^{\frac{3}{2}}}
\\
\\
f'(0) =& \frac{ 1}{2(4 - 0)^{\frac{3}{2}}}
\\
\\
f'(0) =& \frac{1}{2 [(4)^{\frac{1}{2}}]^3z}
\\
\\
f'(0) =& \frac{1}{2 (2)^3}
\\
\\
f'(0) =& \frac{1}{2 (8)}
\\
\\
f'(0) =& \frac{1}{16}
\\
\\
L(x) =& \frac{1}{2} + \frac{1}{16} (x - 0)
\\
\\
L(x) =& \frac{1}{2} + \frac{1}{16 }x

\end{aligned}
\end{equation}
$


So

$\displaystyle \frac{1}{\sqrt{4 - x}} \approx \frac{1}{2} + \frac{1}{6} x$

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

$\displaystyle \left| \left( \frac{1}{2} + \frac{1}{16 } x \right) \right| < 0.1$

Equivalently, we could write

$\displaystyle \frac{1}{\sqrt{4 - x}} - 0.1 < \frac{1}{2} + \frac{1}{16} x < \frac{1}{\sqrt{4 - x}} + 0.1$







This says that the Linear Approximation should lie between the curves obtained by shifting the curve $\displaystyle y = \frac{1}{\sqrt{4 - x}}$ upward and downward by $0.1$. The graph shows the tangent line $\displaystyle y = \frac{1}{2} + \frac{1}{16} x$ intersecting the lower curve $\displaystyle y = \frac{1}{\sqrt{4 - x}} - 0.1$ at A and B. We can estimate the $x$-coordinate of A which is $-3.91$ and the $x$-coordinate of B is $2.14$.

Thus, referring to the graph the approximation

$\displaystyle y = \frac{1}{\sqrt{4 - x}} \approx \frac{1}{2} + \frac{1}{16} x$

is accurate to within $0.1$ when $-3.91 < x < 2.14$