Answer:
[tex]x & = \dfrac{1}{2} \ln 2[/tex]
Step-by-step explanation:
Given functions:
[tex]\begin{cases}f(x)=x^2+4x\\g(x)=1+e^{2x}\end{cases}[/tex]
Function composition is an operation that takes two functions and produces a third function.
Therefore, the given composite function fg(x) means to substitute the function g(x) in place of the x in function f(x):
[tex]\begin{aligned}f[g(x)] & = [g(x)]^2 + 4[g(x)]\\& = \left(1+e^{2x}\right)^2+4\left(1+e^{2x}\right)\\& = 1+2e^{2x}+(e^{2x})^2+4+4e^{2x}\\& = e^{4x}+6e^{2x}+5\end{aligned}[/tex]
[tex]\textsf{Let }u=e^{2x}[/tex]
[tex]\implies e^{4x}+6x^{2x}+5=u^2+6u+5[/tex]
Set the composite function to 21 and factor:
[tex]\begin{aligned}f[g(x)] & = 21\\\implies u^2+6u+5 & = 21\\u^2+6u-16 & = 0\\u^2+8u-2u-16 & = 0\\u(u+8)-2(u+8) & = 0\\(u-2)(u+8) & = 0\\\end{aligned}[/tex]
Apply the zero-product property:
[tex]\implies (u-2)=0 \implies u=2[/tex]
[tex]\implies (u+8)=0 \implies u=-8[/tex]
[tex]\textsf{Substitute back in }u=e^{2x}:[/tex]
[tex]\implies e^{2x}=2[/tex]
[tex]\implies e^{2x}=-8[/tex]
As we cannot take logs of negative numbers, the only solution is:
[tex]\begin{aligned}e^{2x} & = 2\\\ln e^{2x} & = \ln 2\\2x \ln e & = \ln 2\\2x(1) & = \ln 2\\2x & = \ln 2\\\implies x & = \dfrac{1}{2} \ln 2\end{aligned}[/tex]
Learn more about composite functions here:
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