Find two functions f(x) and g(x) so that h(x)=(f∘g)(x) a) h(x)=(2x+3)2 b) h(x)=|3−2x−x2|. c) h(x)=h(x)=53−x...

Given data,
The function ${\displaystyle h\left(x\right)}$ is defined as ${\displaystyle h\left(x\right)=fog\left(x\right)}$
Step 1
a)Here,
${\displaystyle h\left(x\right)={(2x+3)}^2}$
It is given that,
${\displaystyle h\left(x\right)=fog\left(x\right)}$
By comparing both equations,
fοg ${\displaystyle \left(x\right)={(2x+3)}^2}$
${\displaystyle f\left(x\right)\times g\left(x\right)=(2x+3)\times (2x+3)}$

b)Here,
${\displaystyle h\left(x\right)=\left|3-2x-x^2\right|}$
It is given that,
${\displaystyle h\left(x\right)=f\text{οg}\left(x\right)}$
By comparing both equations,
fοg(x)f(x)×g(x)f(x)×g(x)Hence,
${\displaystyle \text{fog}\left(x\right)=\left|3-2x-x^2\right|}$
${\displaystyle f\left(x\right)\times g\left(x\right)=\left|3-3x+x-x^2\right|}$
${\displaystyle f\left(x\right)\times g\left(x\right)=\left|(-x+1)(x+3)\right|}$
Hence, ${\displaystyle f\left(x\right)=(1-x)}$ and ${\displaystyle g\left(x\right)=(x+3)}$

Step 2
c)Here,
$h(x)=\frac{5}{\sqrt{3-x}}$
It is given that,
$h(x)=f\text{οg}(x)$
By comparing both equations,
$f\text{og}=\frac{5}{\sqrt{3-x}}$
$f(x)\times g(x)=5\times \frac{1}{\sqrt{3-x}}$
Hence, $f(x)=5$ and $g(x)=\frac{1}{\sqrt{3-x}}$
Step 3
d)Here,
$f(x)=\sqrt[5]{x^3+4x^2+1}$
It is given that,
$h(x)=f\text{οg}(x)$
By comparing both equations,
$f\text{og}(x)=\sqrt[5]{x^3+4x^2+1}$
$f(x)\times g(x)=\sqrt[2]{x^3+4x^2+1}\times \sqrt[3]{x^3+4x^2+1}$
Hence, $f(x)=\sqrt[2]{x^3+4x^2+1}andg(x)=\sqrt[3]{x^3+4x^2+1}$