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Rudiy27
3 years ago
5

Please help!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
elena55 [62]3 years ago
8 0

h(x)=(f\circ g)(x)\\\\h(x)=\sqrt[3]{x+3};\ f(x)=\sqrt[3]{x+2}\\\\h(x)=\sqrt[3]{x+1+2}=\sqrt[3]{(x+1)+2}=\sqrt[3]{g(x)+2}\to \boxed{g(x)=x+1}

-----------------------------------------------------------------------------------------------

f(x)=2x;\ g(x)=2x-1;\ h(x)=\sqrt{x}\\\\(f\circ g\circ h)(9)=?\\\\h(9)=\sqrt9=3\\\\(g\circ h)(9)=g(3)=2(3)-1=6-1=5\\\\(f\circ g\circ h)(9)=f(5)=2(5)=10\\------------------------------\\other\ method:\\\\(f\circ g\circ h)(x)=2(2\sqrt{x}-1)=4\sqrt{x}-2\\\\(f\circ g\circ h)(9)=4\sqrt9-2=4(3)-2=12-2=10

-------------------------------------------------------------------------------------------------

f(x)=4x+1;\ g(x)=x^2-5\\\\(f\circ g)(4)=?\\\\g(4)=4^2-5=16-5=11\\(f\circ g)(4)=f(11)=4(11)+1=44+1=45\\----------------------------------\\other\ method:\\\\(f\circ g)(x)=4(x^2-5)+1=4x^2-20+1=4x^2-19\\\\(f\circ g)(4)=4(4)^2-19=4(16)-19=64-19=45


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Since the equation is undefined for -2

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\mathrm{Find\:Least\:Common\:Multiplier\:of\:}a+2,\:-4+a^2,\:a-2:\quad \left(a+2\right)\left(a-2\right)

\mathrm{Multiply\:by\:LCM=}\left(a+2\right)\left(a-2\right)

\frac{3}{a+2}\left(a+2\right)\left(a-2\right)-\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right)=\frac{1}{a-2}\left(a+2\right)\left(a-2\right)

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  • -\frac{6a}{-4+a^2}\left(a+2\right)\left(a-2\right):\quad -6a
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\mathrm{Divide\:both\:sides\:by\:}-4

\frac{-4a}{-4}=\frac{8}{-4}

a=-2

\mathrm{Verify\:Solutions}

\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{3}{a+2}-6\frac{a}{-4+a^2}-\frac{1}{a-2}\mathrm{\:and\:compare\:to\:zero}

\mathrm{Solve\:}\:a+2=0:\quad a=-2

\mathrm{Solve\:}\:-4+a^2=0:\quad a=2,\:a=-2

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a=-2,\:a=2

Since the equation is undefined for -2

Therefore, NO SOLUTION for the given equation.

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