$f:\mathbb R \to (0,\infty)$ defined by $f(x)=e^x$. Describe its inverse.

Question

How do I go about describing it? Well first is the inverse $e^{-x}$ or $\ln(x)$?

Additionally, since I have no clue how to solve these problems as I am probably overthinking them...

$f:\mathbb R\to \mathbb R$ defined by $f(x) = x^3 + 1$. Describe its inverse.

Is the inverse of this function $\sqrt[3]{x-1}$? Is that what they mean by describing it?

Answer 1

The inverse of a function is with respect to function composition, not with respect to point-wise addition. So, the inverse of $e^x$ is not $x^{-1}$ but rather $\ln (x)$. Can you now work out the second function?

As for describing these functions, it's a bit unclear what they aim at. Perhaps to draw a sketch of the graph, or to say something qualitative about it by looking at the derivative (it's sign etc.). Perhaps they want you to relate the qualitative properties of the inverse from the original function. Who knows.

Answer 2

The inverse of a function $\mathrm{f}$ is the function, denoted by $\mathrm{f}^{-1}$, for which $$(\mathrm{f}\circ \mathrm{f}^{-1})(x) \equiv x \equiv (\mathrm{f}^{-1}\circ\mathrm{f})(x)$$ where $(\mathrm{a}\circ \mathrm{b})(x)$ means $\mathrm{a}(\mathrm{b}(x))$ for all $x$, i.e. substitute $x=\mathrm{b}(x)$ into $\mathrm{a}(x)$.

Let $\mathrm{f}(x) = \mathrm{e}^x$, $\mathrm{g}(x) = \mathrm{e}^{-x}$ and $\mathrm{h}(x)=\ln x$. We have \begin{eqnarray*} (\mathrm{f}\circ \mathrm{g})(x) &=& \mathrm{e}^{\mathrm{e}^{-x}} \not\equiv x\\ \\ (\mathrm{f}\circ \mathrm{h})(x) &=& \mathrm{e}^{\ln x} \equiv x \end{eqnarray*} Likewise $(\mathrm{h}\circ\mathrm{f})(x) = \ln \mathrm{e}^x \equiv x$. Clearly $\mathrm{f}^{-1}(x) = \ln x$. The inverse of $x \mapsto \mathrm{e}^x$ is $x \mapsto \ln x$.

Answer 3

This answer is quite a 'heavy' one. Only pay attention to it if it really helps you.

Let $X$ and $Y$ be sets. Then a function $f:X\rightarrow Y$ can be described as a triple $\left(X,G,Y\right)$ where $G$ is a subset of $X\times Y$ having the special property that for every $x\in X$ there is exactly one $y\in Y$ such that $\left(x,y\right)\in G$. This $y$ is denoted as $f\left(x\right)$. The function has an inverse $f^{-1}:Y\rightarrow X$ if and only if for each $y\in Y$ there is exactly one $x\in X$ such that $\left(x,y\right)\in G$. This $x$ is then denoted as $f^{-1}\left(y\right)$ and $f^{-1}$ is the triple $\left(Y,G^{op},X\right)$ where $G^{op}$ is a subset of $Y\times X$ defined by $G^{op}=\left\{ \left(y,x\right)\mid\left(x,y\right)\in G\right\} $.

Characteristic are the equalities: $$f^{-1}\left(f\left(x\right)\right)=x$$ for each $x\in X$ and: $$f\left(f^{-1}\left(y\right)\right)=y$$ for each $y\in Y$.

In your case we deal with $X=\mathbb{R}$ and $Y=\left(0,\infty\right)$ and $G=\left\{ \left(x,e^{x}\right)\mid x\in\mathbb{R}\right\} $.

The inverse of $f=\left(\mathbb{R},G,\left(0,\infty\right)\right)$ is $f^{-1}=\left(\left(0,\infty\right),G^{op},\mathbb{R}\right)$ where $G^{op}=\left\{ \left(y,\ln\left(y\right),\right)\mid y\in\left(0,\infty\right)\right\} $.

Note that indeed $\ln (e^{x})=x$ for each $x\in\mathbb{R}$ and $e^{\ln y}=y$ for each $y\in\left(0,\infty\right)$.

Answer 4

Well inverse of $e^x$ is $\ln(x)$ which is monotonically increasing continuous smooth infinitely differentiable concave function which has one zero which is 1.Inverse of $x^3+1$ is $\sqrt[3]{x-1}$ and is defined for $x\geq 1$,has one zero at $x=1$ and is monotonically increasing continuous smooth infinitely differentiable function