Answer
(a) $L(x) = x+1$
$\sqrt[3] {1.03}\approx 1.01$
(b) We can see a sketch of the graph below.
The linear approximation is accurate to within $0.1$ for the values $-0.2 \leq x \leq 0.4$
Work Step by Step
(a) $f(x) = \sqrt[3] {1+3x}$
$f'(x) = \frac{1}{3}(1+3x)^{-2/3}~(3)$
$f'(x) = \frac{1}{(1+3x)^{2/3}}$
When $x = 0$:
$f(0) = \sqrt[3] {1+3(0)} = 1$
$f'(0) = \frac{1}{(1+3(0))^{2/3}} = 1$
We can find the linearization near $a=0$:
$L(x) = f(a)+f'(a)(x-a)$
$L(x) = f(0)+f'(0)(x-0)$
$L(x) = 1+1(x)$
$L(x) = x+1$
We can find an approximate value for $\sqrt[3] {1.03}$:
$\sqrt[3] {1.03}=\sqrt[3] {1+3(0.01)}\approx 1+0.01=1.01$
(b) We can see a sketch of the graph below.
With the help of the graph, we can calculate that the linear approximation is accurate to within $0.1$ for the values $-0.2 \leq x \leq 0.4$
We can verify this:
When $x = -0.2$:
$x+1 = (-0.2)+1 = 0.8$
$f(-0.2) = \sqrt[3] {1+3(-0.2)} = 0.74$
When $x = 0.4$:
$x+1 = (0.4)+1 = 1.4$
$f(0.4) = \sqrt[3] {1+3(0.4)} = 1.3006$
