If g(x)=3 x^2+2 x-3, f(0)=-3 and 4 g(f(x))=3 x^2-32 x+72, then f(g(2)) is equal to:

Question

If $g(x)=3 x^2+2 x-3, f(0)=-3$ and $4 g(f(x))=3 x^2-32 x+72$, then $f(g(2))$ is equal to:

Accepted Answer

Given

$g(x)=3{x}^{2}+2x-3,f(0)=-3$, and $4g(f(x))=3{x}^{2}-32x+72$

Since $g(x)$ is a quadratic function (degree 2) and the composite function $4g(f(x))$ is also a quadratic expression $(3{x}^{2}-32x+72)$, $f(x)$ must be a linear function.

Let $f(x)=ax+b$.

Using $f(0)=-3:a(0)+b=-3\implies b=-3$.

So, $f(x)=ax-3$.

First, find $g(f(x))$ by substituting $f(x)$ into $g(x)$ :

$g(f(x))=3(f(x){)}^{2}+2f(x)-3$

Substitute this into the given equation $4g(f(x))=3{x}^{2}-32x+72$ :

$4[3(ax-3{)}^{2}+2(ax-3)-3]=3{x}^{2}-32x+72$

Solve for the Coefficient ' $a$ '

Expand the left side of the equation:

$4[3({a}^{2}{x}^{2}-6ax+9)+2ax-6-3]=3{x}^{2}-32x+72$

$4[3{a}^{2}{x}^{2}-18ax+27+2ax-9]=3{x}^{2}-32x+72$

$4[3{a}^{2}{x}^{2}-16ax+18]=3{x}^{2}-32x+72$

$12{a}^{2}{x}^{2}-64ax+72=3{x}^{2}-32x+72$

Compare the coefficients:

For ${x}^{2}:12{a}^{2}=3\implies {a}^{2}=\frac{1}{4}\implies a=\pm \frac{1}{2}$

For $x$ : $-64a=-32\implies a=\frac{32}{64}=\frac{1}{2}$ Thus, $f(x)=\frac{1}{2}x-3$. Calculate $f(g(2))$

Find $g(2)$

$g(2)=3(2{)}^{2}+2(2)-3=12+4-3=13$

Step B: Find $f(g(2))$ which is $f(13): B: Find $f(g(2))$ which is $f(13)$

$$$\begin{gathered} f(13)=\frac{1}{2}(13)-3=6.5-3=3.5=\frac{7}{2} \end{gathered}$$

Solutions · 1