Answer
$(1+x)^{-3} \approx 1-3x$
On the graph, we can see that the linear approximation is accurate to within 0.1 when $-0.1 \leq x \leq 0.1$
Work Step by Step
$f(x) = (1+x)^{-3}$
$f'(x) = -3~(1+x)^{-4}$
When $x = 0$:
$f'(x) = -3~(1+x)^{-4}$
$f'(0) = -3~(1+0)^{-4}$
$f'(0) = -3$
We can find the linear approximation at $a=0$:
$f(x) \approx f(a)+f'(a)(x-a)$
$f(x) \approx f(0)+f'(0)(x-0)$
$f(x) \approx 1+(-3)(x)$
$f(x) \approx 1-3x$
$(1+x)^{-3} \approx 1-3x$
On the graph, we can see that the linear approximation is accurate to within 0.1 when $-0.1 \leq x \leq 0.1$
We can verify this:
When $x = -0.1$:
$1-3x = 1-(3)(-0.1) = 1.3$
$(1+x)^{-3} = (1-0.1)^{-3} = 1.37$
When $x = 0.1$:
$1-3x = 1-(3)(0.1) = 0.7$
$(1+x)^{-3} = (1+0.1)^{-3} = 0.75$
