Answer
$\sqrt[4] {1+2x} \approx 1+\frac{1}{2}~x$
On the graph, we can see that the linear approximation is accurate to within 0.1 when $-0.3 \leq x \leq 0.6$
Work Step by Step
$f(x) = \sqrt[4] {1+2x}$
$f'(x) = \frac{1}{4}~(1+2x)^{-3/4}\cdot (2)$
$f'(x) = \frac{1}{2~(1+2x)^{3/4}}$
When $x = 0$:
$f'(x) = \frac{1}{2~(1+2x)^{3/4}}$
$f'(0) = \frac{1}{2~(1+2(0))^{3/4}}$
$f'(0) = \frac{1}{2}$
We can find the linear approximation at $a=0$:
$f(x) \approx f(a)+f'(a)(x-a)$
$f(x) \approx f(0)+f'(0)(x-0)$
$f(x) \approx 1+\frac{1}{2}~x$
$\sqrt[4] {1+2x} \approx 1+\frac{1}{2}~x$
On the graph, we can see that the linear approximation is accurate to within 0.1 when $-0.3 \leq x \leq 0.6$
We can verify this:
When $x = -0.3$:
$1+\frac{1}{2}~x = 1+\frac{1}{2}~(-0.3) = 0.85$
$\sqrt[4] {1+2x} = \sqrt[4] {1+2(-0.3)} = 0.795$
When $x = 0.6$:
$1+\frac{1}{2}~x = 1+\frac{1}{2}~(0.6) = 1.30$
$\sqrt[4] {1+2x} = \sqrt[4] {1+2(0.6)} = 1.22$
