# If f(x)=2x^3/5+7 ,find f^-1(x)?

Jun 21, 2018

${f}^{- 1} \left(x\right) = \setminus \sqrt[3]{\frac{5}{2} \left(x - 7\right)}$

#### Explanation:

Given a function $y = f \left(x\right)$, you can find its inverse $x = {f}^{- 1} \left(y\right)$ by solving the equation for $x$. By doing so, you change the roles of dependant and independant variables.

So, if we start from

$y = \frac{2}{5} {x}^{3} + 7$

We subtract $7$ from both sides:

$y - 7 = \frac{2}{5} {x}^{3}$

We multiply both sides by $\frac{5}{2}$:

$\frac{5}{2} \left(y - 7\right) = {x}^{3}$

And finally extract the third root:

$\setminus \sqrt[3]{\frac{5}{2} \left(y - 7\right)} = x$

So, given $f \left(x\right) = \frac{2}{5} {x}^{3} + 7$, we have ${f}^{- 1} \left(x\right) = \setminus \sqrt[3]{\frac{5}{2} \left(x - 7\right)}$

As you can see here, the graphs of a function and its inverse are symmetrical with respect to $y = x$.