The function $\mathrm{f}$ is defined by $\mathrm{f}\left( x \right)=\left\{ \begin{matrix}{{x}^{2}}+

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}\left( x ight)=\left\{ \begin{matrix}{{x}^{2}}+1, & \mathrm{for }x\ge 0\ 2x+1, & \mathrm{for }xSketch the graph of $\mathrm{f}$. Show that ${{\mathrm{f}}^{-1}}$ exists and define it. Sketch the graph of ${{\mathrm{f}}^{-1}}$ on the same diagram as the graph of $\mathrm{f}$, and find the value of $x$ for which $\mathrm{f}\left( x ight)={{\mathrm{f}}^{-1}}\left( x ight)$. Given $\mathrm{g}\left( x ight)={{x}^{2}}-2$, $x\in \mathbb{R}$, find $\mathrm{gf}\left( x ight)$.

Accepted Answer

Given $x=-1$, $	herefore \mathrm{gf}\left( x ight)=\left\{ \begin{matrix}{{\left( {{x}^{2}}+1 ight)}^{2}}-2, & \mathrm{for }\!\!~\!\!\mathrm{ }\,\,\,x\ge 0 \ {{\left( 2x+1 ight)}^{2}}-2, & \mathrm{for }\!\!~\!\!\mathrm{ }\,\,\,x<0 \\end{matrix} ight.$:

EJC Functions Tutorial Q11

EJC Functions Tutorial Q11

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