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# If the function $f : R \rightarrow R$ is given by $f(x) =\large \frac{x+3}{2}$ and $g : R \rightarrow R$ is given by $g(x)=2x-3,$ find (i) fog and (ii) gof. Is $f^{-1}=g$?

Can you answer this question?

Step 1:
$f(x)=\large\frac{x+3}{2}$
$g(x)=2x-3$
$fog=\large\frac{2x-3+3}{2}$
Step 2:
$gof=2\big(\large\frac{x+3}{2}\big)$$-3 \qquad=x \therefore fog=gof Step 3: Now let us find f^{-1} Let f(x)=y \Rightarrow \large\frac{x+3}{2}=$$y$
$x+3=2y$
$x=2y-3$
Clearly $2y-3\in R$ for all $y\in R$
Step 4:
Thus for all $y\in R$ there exists $x=\large\frac{y+7}{3}$$\in R$
Such that $f(x)=f(2y-3)$
$\Rightarrow \large\frac{2y-3+3}{2}$
$\Rightarrow y$
$\therefore f^{-1}(y)=2y-3$
$\therefore f^{-1}:R\rightarrow R$ is given by
$f^{-1}(x)=2x-3$
Which is $g(x)$
$\therefore f^{-1}(x)=g(x)$
answered Oct 1, 2013