Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 6

Determine the Linear Approximation of the function $g(x) = \sqrt{1 + x}$ at $a = 0$ and use it to approximate the numbers $\sqrt[3]{0.95}$ and $\sqrt[3]{1.1}$. Illustrate by graphing $g$ and the tangent line.

Using the Linear Approximation/Tangent Line Approximation

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


$
\begin{equation}
\begin{aligned}

f(a) = f(0) =& \sqrt[3]{1 + 0}
\\
\\
f(0) =& \sqrt[3]{1}
\\
\\
f(0) =& 1
\\
\\
f'(a) = f'(0) =& \frac{d}{dx} (\sqrt[3]{\sqrt{1 + x}})
\\
\\
f'(0) =& \frac{d}{dx} (1 + x)^{\frac{1}{3}}
\\
\\
f'(0) =& \frac{1}{3} (1 + x)^{\frac{-2}{3}}
\\
\\
f'(0) =& \frac{1}{3 (1 + x)^{\frac{2}{3}}}
\\
\\
f'(0) =& \frac{1}{3 (1 + 0)^{\frac{2}{3}}}
\\
\\
f'(0) =& \frac{1}{3(1)^{\frac{2}{3}}}
\\
\\
f'(0) =& \frac{1}{3}
\\
\\
L(x) =& 1 + \frac{1}{3} (x - 0)
\\
\\
L(x) =& 1 + \frac{1}{3} x

\end{aligned}
\end{equation}
$


For $\sqrt[3]{0.95}$


$
\begin{equation}
\begin{aligned}

\sqrt[3]{0.95} =& \sqrt[3]{1 + x}
\\
\\
0.95 =& 1 + x
\\
\\
x =& 0.95 - 1
\\
\\
x =& -0.05

\end{aligned}
\end{equation}
$


For $\sqrt{\sqrt[3]{1.1}}$


$
\begin{equation}
\begin{aligned}

\sqrt[3]{1.1} = \sqrt[3]{1 + x}
\\
\\
1.1 =& 1 + x
\\
\\
x =& 1.1 - 1
\\
\\
x =& 0.1

\end{aligned}
\end{equation}
$


Linear Approximation gives


$
\begin{equation}
\begin{aligned}

L(-0.05) =& 1 + \frac{1}{3} (-0.05)
\\
\\
L(-0.05) =& \frac{3 + (-0.05)}{3}
\\
\\
L(-0.05) =& \frac{3 - 0.05}{3}
\\
\\
L(-0.05) =& \frac{2.95}{3}
\\
\\
L(-0.05) =& 0.983

\end{aligned}
\end{equation}
$


and


$
\begin{equation}
\begin{aligned}

L(0.1) =& 1 + \frac{1}{3} (0.1)
\\
\\
L(0.1) =& \frac{3 + 0.1}{3}
\\
\\
L(0.1) =& \frac{3.1}{3}
\\
\\
L(0.1) =& 1.033

\end{aligned}
\end{equation}
$