Answer
$g(x) \approx 1+\frac{x}{3}$
$\sqrt[3] {0.95} \approx 0.983$
$\sqrt[3] {1.1} \approx 1.033$
We can see a sketch of the graphs below.
Work Step by Step
$g(x) = \sqrt[3] {1+x}$
$g'(x) = \frac{1}{3~(1-x)^{2/3}}$
When $x = 0$:
$g'(x) = \frac{1}{3~(1-x)^{2/3}}$
$g'(0) = \frac{1}{3~(1-0)^{2/3}}$
$g'(0) = \frac{1}{3}$
We can find the linear approximation at $a=0$:
$g(x) \approx g(a)+g'(a)(x-a)$
$g(x) \approx g(0)+g'(0)(x-0)$
$g(x) \approx 1+(\frac{1}{3})(x)$
$g(x) \approx 1+\frac{x}{3}$
We can approximate $\sqrt[3] {0.95} = g(-0.05)$:
$g(x) \approx 1+\frac{x}{3}$
$g(-0.05) \approx 1+\frac{-0.05}{3}$
$g(-0.05) \approx 0.983$
$\sqrt[3] {0.95} \approx 0.983$
We can approximate $\sqrt[3] {1.1} = g(0.1)$:
$g(x) \approx 1+\frac{x}{3}$
$g(0.1) \approx 1+\frac{0.1}{3}$
$g(0.1) \approx 1.033$
$\sqrt[3] {1.1} \approx 1.033$
We can see a sketch of the graphs below.