Frank Guyton

Answered

2021-12-28

Find two functions $f\left(x\right)$ and $g\left(x\right)$ so that $h\left(x\right)=\left(f\circ g\right)\left(x\right)$
a) $h\left(x\right)={\left(2x+3\right)}^{2}$
b) $h\left(x\right)=|3-2x-{x}^{2}|$.
c) $h\left(x\right)=h\left(x\right)=\frac{5}{\sqrt{3-x}}$
d) $h\left(x\right)=\sqrt{5}\left\{{x}^{3}+4{x}^{2}+1\right\}$

Answer & Explanation

Buck Henry

Expert

2021-12-29Added 33 answers

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

kaluitagf

Expert

2021-12-30Added 38 answers

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

karton

Expert

2022-01-09Added 439 answers

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

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