# Let f(x)=x^5+3x−2 and let f^-1 denote the inverse of f, Then how do you find (f^-1)'(2)?

Nov 21, 2016

$\left[{f}^{-} 1\right] ' \left(2\right) = \frac{1}{8}$

#### Explanation:

For the sake of easier notation, we shall say that $f$ is a function with inverse function $g$, that is, ${f}^{-} 1 \left(x\right) = g \left(x\right)$.

According to the definition of inverse functions,

$f \left(g \left(x\right)\right) = x$

Differentiating through the chain rule gives

$f ' \left(g \left(x\right)\right) g ' \left(x\right) = 1$

Solving for the derivative of the inverse gives

$g ' \left(x\right) = \frac{1}{f ' \left(g \left(x\right)\right)}$

So, we want to find $g ' \left(2\right)$.

g'(2)=1/(f'(g(2))

We want to first find $g \left(2\right)$, however, we cannot write an expression for $g \left(x\right)$. What we have to remember is that the domain of the mother function is the range of its inverse function, and vice versa.

Note that if $f \left(x\right) = 2$, then $x = g \left(2\right)$. This means we should let $f \left(x\right) = 2$ then solve for $x$, which is equal to $g \left(2\right)$.

$f \left(x\right) = 2 \text{ "=>" } {x}^{5} + 3 x - 2 = 2$

Continuing to solve yields

${x}^{5} + 3 x - 4 = 0$

This may look impossible, but notice that the sum of the coefficients of each term is $0$, that is, $1 + 3 - 4 = 0$. This means that $x = 1$ is a solution.

Dividing, we see that

$\left(x - 1\right) \left({x}^{4} + {x}^{3} + {x}^{2} + x + 4\right) = 0$

Note that ${x}^{4} + {x}^{3} + {x}^{2} + x + 4 > 0$ for all values of $x$, so it has no real roots. Thus, $f \left(x\right) = 2$ when $x = 1$, i.e., $f \left(1\right) = 2$.

This means that $g \left(2\right) = 1$.

Returning to what we had earlier, we see that

$g ' \left(2\right) = \frac{1}{f ' \left(g \left(2\right)\right)} = \frac{1}{f ' \left(1\right)}$

Find this by taking the derivative of $f$.

$f \left(x\right) = {x}^{5} + 3 x - 2 \text{ "=>" } f ' \left(x\right) = 5 {x}^{4} + 3$

Thus $f ' \left(1\right) = 5 + 3 = 8$.

$g ' \left(2\right) = \frac{1}{8}$

$\left[{f}^{-} 1\right] ' \left(2\right) = \frac{1}{8}$