Question

Let f:R to R be defined by f(x) = 3x²-5 and g:R to R be defined as g(x)=3x, find gof(-2)​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:f : R \: \to \: R$

$\rm :\longmapsto\:g : R \: \to \: R$

defined as

$\rm :\longmapsto\:f(x) = {3x}^{2} - 5$

$\rm :\longmapsto\:g(x) = 3x$

$\purple{\large\underline{\sf{To\:Find - }}}$

$\rm :\longmapsto\:gof( - 2)$

$\red{\large\underline{\sf{Solution-}}}$

Given that,

$\rm :\longmapsto\:f : R \: \to \: R$

defined as

$\rm :\longmapsto\:f(x) = {3x}^{2} - 5$

and

$\rm :\longmapsto\:g : R \: \to \: R$

defined as

$\rm :\longmapsto\:g(x) = 3x$

Now, Consider

$\rm :\longmapsto\:gof( - 2)$

$\rm \: = \: g[f( - 2)]$

$\rm \: = \: g\bigg[3 {( - 2)}^{2} - 5\bigg]$

$\rm \: = \: g\bigg[3 \times 4 - 5\bigg]$

$\rm \: = \: g\bigg[12 - 5\bigg]$

$\rm \: = \: g\bigg[7\bigg]$

$\rm \: = \: 3 \times 7$

$\rm \: = \: 21$

Hence,

$\sf\implies \: \boxed{\tt{ \:gof( - 2) \: = \: 21 \: }}$

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MORE TO KNOW

1. Modulus function : is defined as

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{ - x \: \: when \: x \: < \: 0} \\ \\ &\sf{ \: \:x \: \: \:when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{Domain \: = \: ( - \infty , \infty )} \\ \\ &\sf{Range \: [0, \infty )} \end{cases}\end{gathered}\end{gathered}$

2. Signum Function :- is defined as

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: sgn(x) = \dfrac{ |x| }{x} = \begin{cases} &\sf{ - 1 \: \: when \: x \: < \: 0}\\ \\ &\sf{ \: \:0 \: \: \:when \: x = 0} \\ \\ &\sf{ \: \:1 \: \: \:when \: x > 0} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{Domain \: = \: ( - \infty , \infty )} \\ \\ &\sf{Range \: \{ - 1,0,1 \}} \end{cases}\end{gathered}\end{gathered}$