# Let f(x)= -35x-x^5 and let g be the inverse function of f, how do you find a) g(0) b) g'(0) c) g(-36) d) g'(-36)?

Aug 4, 2017

Parts a, c by inspection. Parts b, d use g'(a) = 1/(f'(g(a))

#### Explanation:

For constant $a$, $g \left(a\right)$ is the solution to $f \left(x\right) = a$

Parts a, b
$g \left(0\right) = 0$ because $f \left(0\right) = 0$

Note that $f ' \left(x\right) = - 35 - 5 {x}^{4}$, so $f ' \left(0\right) = - 35$ and

$g ' \left(0\right) = \frac{1}{f ' \left(g \left(0\right)\right)} = \frac{1}{f ' \left(0\right)} = \frac{1}{- 35} = - \frac{1}{35}$

Parts c, d
$g \left(- 36\right) = 1$ because $f \left(1\right) = - 36$

Note that $f ' \left(x\right) = - 35 - 5 {x}^{4}$, so $f ' \left(1\right) = - 40$ and

$g ' \left(- 36\right) = \frac{1}{f ' \left(g \left(- 36\right)\right)} = \frac{1}{f ' \left(1\right)} = \frac{1}{- 40} = - \frac{1}{40}$