Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 4

Determine the Linearization $L(x)$ of the function $\displaystyle f(x) = x^{\frac{3}{4}}$ at $\displaystyle a = 16$.

Using the Linearization Formula

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


$
\begin{equation}
\begin{aligned}

f(a) = f(16) =& (16)^{\frac{3}{4}}
\\
\\
f(16) =& [(16)^{\frac{1}{4}}]^3
\\
\\
f(16) =& (2)^3
\\
\\
f(16) =& 8
\\
\\
f'(a) f'(16) =& \frac{d}{dx} (x)^{\frac{3}{4}}
\\
\\
f'(16) =& \frac{3}{4} (x)^{\frac{-1}{4}}
\\
\\
f'(16) =& \frac{3}{4(x)^{\frac{1}{4}}}
\\
\\
f'(16) =& \frac{3}{4 (16)^{\frac{1}{4}}}
\\
\\
f'(16) =& \frac{3}{(4)(2)}
\\
\\
f'(16) =& \frac{3}{8}
\\
\\
L(x) =& 8 + \frac{3}{8} (x - 16)
\\
\\
L(x) =& 8 + \frac{3}{8} x 0 \frac{(3)(16)}{8}
\\
\\
L(x) =& 8 + \frac{3}{8} x - (3)(2)
\\
\\
L(x) =& 8 + \frac{3}{8} x - 6
\\
\\
L(x) =& \frac{3}{8} x + 2

\end{aligned}
\end{equation}
$