Answer
At $a=0,~~~$ $ln~(1+x) \approx x$
On the graph, we can see that the linear approximation is accurate to within 0.1 when $-0.3 \leq x \leq 0.5$
Work Step by Step
$f(x) = ln~ (1+x)$
$f'(x) = \frac{1}{1+x}$
When $x = 0$:
$f'(x) = \frac{1}{1+x}$
$f'(x) = \frac{1}{1+0}$
$f'(0) = 1$
We can find the linear approximation at $a=0$:
$f(x) \approx f(a)+f'(a)(x-a)$
$f(x) \approx f(0)+f'(0)(x-0)$
$f(x) \approx ln~(1)+(1)(x)$
$f(x) \approx 0+x$
$f(x) \approx x$
$ln~(1+x) \approx x$
On the graph, we can see that the linear approximation is accurate to within 0.1 when $-0.3 \leq x \leq 0.5$
We can verify this:
When $x = -0.3$:
$x = -0.3$
$ln~(1+x) = ln~(1-0.3) = -0.36$
When $x = 0.5$:
$x = 0.5$
$ln~(1+x) = ln~(1+0.5) = 0.405$
