Practice section 1 of 2

State and justify which of the following is a one-to-one function. (a) $f:x\mapsto\ln x-x$, $x\in\mathbb R$, $1<x\le e^2$. (b) $g:x\mapsto(x+2)^2+e^x$, $x\in\mathbb R$, $x>-4$.

Given that $f:x\mapsto x^2$, $x\in\mathbb R$, $-2\le x\le3$. Show that $f$ is not one-to-one. A function $g$ is a restriction on $f$; state the largest possible domain of $g$, for which $g$ is one-to-one.

Let $f$ be a function defined by $f:x\mapsto2x+1$, $x\in\mathbb R$, $x>0$. Find $f^{-1}(x)$.

Let $f$ be the function defined by $f:x\mapsto x^2-1$, $x\in\mathbb R$, $x<0$. Define $f^{-1}$ in a similar form.

Let $f$ be a function defined by $f:x\mapsto e^x$, $x\in\mathbb R$. (i) Define $f^{-1}$ in a similar form. (ii) On a single diagram, sketch $y=f(x)$, $y=f^{-1}(x)$ and $y=x$.

Determine the range of each of the following functions. State, with reasons, whether the inverse function exists in each case. Then, find the rule and domain of the inverse function if it exists.

(a) $h:x\mapsto\cos(x+1)$, $-1\leq x\leq-1+\frac{\pi}{2}$.

(b) $q:x\mapsto e^x(x-1)$, $x\in\mathbb R$.