# What is the inverse of y=3ln(5x)+x^3 ?

May 14, 2016

${f}^{- 1} \left(y\right) = x : f \left(x\right) = y$

#### Explanation:

Let $f \left(x\right) = 3 \ln \left(5 x\right) + {x}^{3}$

Let us assume that we are dealing with Real values and therefore the Real natural logarithm.

Then we are constrained to $x > 0$ in order that $\ln \left(5 x\right)$ be defined.

For any $x > 0$ both terms are well defined and so $f \left(x\right)$ is a well defined function with domain $\left(0 , \infty\right)$.

Note that $3 \ln \left(5\right)$ and ${x}^{3}$ are both strictly monotonic increasing on this domain so our function is too and is one-to-one.

For small positive values of $x$, the term ${x}^{3}$ is small and positive and the term $3 \ln \left(5 x\right)$ is arbitrarily large and negative.

For large positive values of $x$, the term $3 \ln \left(5 x\right)$ is positive and the term ${x}^{3}$ is arbitrarily large and positive.

Since the function is also continuous, the range is $\left(- \infty , \infty\right)$

So for any value of $y \in \left(- \infty , \infty\right)$ there is a unique value of $x \in \left(0 , \infty\right)$ such that $f \left(x\right) = y$.

This defines our inverse function:

${f}^{- 1} \left(y\right) = x : f \left(x\right) = y$

That is ${f}^{- 1} \left(y\right)$ is the value of $x$ such that $f \left(x\right) = y$.

We have shown (informally) that this exists, but there is no algebraic solution for $x$ in terms of $y$.

The graph of ${f}^{- 1} \left(y\right)$ is the graph of $f \left(x\right)$ reflected in the line $y = x$.

In set notation:

$f = \left\{\left(x , y\right) \in \left(0 , \infty\right) \times \mathbb{R} : y = 3 \ln \left(5 x\right) + {x}^{3}\right\}$

${f}^{- 1} = \left\{\left(x , y\right) \in \mathbb{R} \times \left(0 , \infty\right) : x = 3 \ln \left(5 y\right) + {y}^{3}\right\}$