## Calculus: Early Transcendentals 8th Edition

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 $-0.3 \leq x \leq 0.5$
$f(x) = ln~ (1+x)$ $f'(x) = \frac{1}{1+x}$ 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 $x = 0.5$: $x = 0.5$ $ln~(1+x) = ln~(1+0.5) = 0.405$