# How do you determine if the function is a one-to-one function and find the formula of the inverse given f(x) = 5x^3 - 7?

Apr 19, 2018

See below

#### Explanation:

Let $f : \mathbb{R} \to \mathbb{R} , f \left(x\right) = 5 {x}^{3} - 7$ be a function from $\mathbb{R}$ to $\mathbb{R}$.

We want to prove that $f$ is injective over $\mathbb{R}$ and then find its inverse. But we will do it the other way around; ie., we will find the inverse of $f$ and this will be sufficient to prove that $f$ is injective.

$f \left(x\right) = 5 {x}^{3} - 7$

so $f \left(x\right) + 7 = 5 {x}^{3}$

so $\frac{f \left(x\right) + 7}{5} = {x}^{3}$

so ${\left(\frac{f \left(x\right) + 7}{5}\right)}^{\frac{1}{3}} = x$

so ${f}^{-} 1 \left(x\right) = {\left(\frac{x + 7}{5}\right)}^{\frac{1}{3}}$

Now we check that $f \left({f}^{-} 1 \left(x\right)\right) = x$ and ${f}^{-} 1 \left(f \left(x\right)\right) = x$

$f \left({f}^{-} 1 \left(x\right)\right) = 5 {\left({\left(\frac{x + 7}{5}\right)}^{\frac{1}{3}}\right)}^{3} - 7 = \cancel{5} \cdot \frac{x + \cancel{7}}{\cancel{5}} - \cancel{7} = x$

${f}^{-} 1 \left(f \left(x\right)\right) = {\left(\frac{5 {x}^{3} - 7 + 7}{5}\right)}^{\frac{1}{3}} = {\left(\frac{5 {x}^{3}}{5}\right)}^{\frac{1}{3}} = x$

So ${f}^{-} 1$ is the inverse of $f$. Now, since $f$ has an inverse, it must be bijective, and so it must be injective.