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solve this composition of functions: given f(x) = x^2-6 and g(x)° 2x^2+4x and h(x)= -x-2 find f(g(h(1)))

solve this composition of functions: given f(x) = x^2-6 and g(x)° 2x^2+4x and h(x-example-1

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SOLUTION

We were given that


\begin{gathered} f\mleft(x\mright)=x^2-6 \\ g(x)=2x^2+4x\text{ and } \\ h\mleft(x\mright)=-x-2 \\ \text{and we are told to find } \\ f\mleft(g\mleft(h\mleft(1\mright)\mright)\mright)​ \end{gathered}

Firstly, we have to find g(h(x)). This means plugging in h(x) into g(x). Doing this we have


\begin{gathered} g(x)=2x^2+4x \\ h\mleft(x\mright)=-x-2 \\ g\mleft(h\mleft(x\mright)\mright)=2(-x-2)^2+4(-x-2) \\ g(h(x))=2\lbrack(-x-2)(-x-2)\rbrack+4(-x-2) \\ g(h(x))=2\lbrack x^2+2x+2x+4\rbrack-4x-8 \\ g(h(x))=2x^2+4x+4x+8-4x-8 \\ \text{collecting like terms } \\ g(h(x))=2x^2+4x+4x-4x+8-8 \\ g(h(x))=2x^2+4x \end{gathered}

Next, we plug in g(h(x)) into f(x), we have


\begin{gathered} g(h(x))=2x^2+4x \\ f\mleft(x\mright)=x^2-6 \\ f\mleft(g\mleft(h\mleft(x\mright)\mright)\mright)=(​2x^2+4x)^2-6 \end{gathered}

No need to expanding that, as this would be time wasting, now let's find f(g(h(1)))​. To do this, we substitute 1 for x in f(g(h(x)))​, we have


\begin{gathered} f(g(h(x)))=(​2x^2+4x)^2-6 \\ f(g(h(1)))=(​2(1)^2+4(1))^2-6 \\ f(g(h(1)))=(​(2*1)^{}+4)^2-6 \\ f(g(h(1)))=(​2^{}+4)^2-6 \\ f(g(h(1)))=(​6)^2-6 \\ f(g(h(1)))=36^{}-6 \\ f\mleft(g\mleft(h\mleft(1\mright)\mright)\mright)=30 \end{gathered}

Hence the answer is 30

User Rboarman
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