Respuesta :
The given functions are:
[tex]\begin{gathered} g(x)=x^2-64 \\ h(x)=8x+512 \end{gathered}[/tex]1. (h/g)(x):
[tex](\frac{h}{g})(x)=\frac{h(x)}{g(x)}[/tex]Then we need to divide h(x) by g(x):
[tex]\frac{8x+512}{x^2-64}[/tex]The factored form of the denominator is:
[tex]\begin{gathered} x^2-64=(x+8)(x-8) \\ \text{Then} \\ \frac{8x+512}{(x+8)(x-8)} \end{gathered}[/tex]We have to use partial fractions by applying the following model:
[tex]\frac{8x+512}{(x+8)(x-8)}=\frac{a_0}{(x+8)}+\frac{a_1}{(x-8)}[/tex]Multiply both sides of the equations by the denominator:
[tex]\frac{(8x+512)(x+8)(x-8)}{(x+8)(x-8)}=\frac{a_0(x+8)(x-8)}{(x+8)}+\frac{a_1(x+8)(x-8)}{(x-8)}[/tex]Now, simplify:
[tex](8x+512)=a_0(x-8)+a_1(x+8)[/tex]To find a0 and a1, substitute the roots of the factors x=-8 and x=8, and solve:
[tex]\begin{gathered} x=8 \\ (8\cdot8+512)=a_0(8-8)+a_1(8+8) \\ 64+512=a_0(0)+a_1(16) \\ 576=a_1(16) \\ a_1=\frac{576}{16}=36 \\ \text{For x=-8:} \\ (8\cdot-8+512)=a_0(-8-8)+a_1(-8+8) \\ -64+512=a_0(-16)+a_1(0) \\ 448=a_0(-16) \\ a_0=\frac{448}{-16} \\ a_0=-28 \end{gathered}[/tex]Now, replace these values and find the equation:
[tex]\begin{gathered} \frac{a_0}{(x+8)}+\frac{a_1}{(x-8)}=\frac{-28_{}}{x+8}+\frac{36_{}}{x-8} \\ (\frac{h}{g})(x)=\frac{-28_{}}{x+8}+\frac{36_{}}{x-8} \end{gathered}[/tex]2. (h*g)(x):
[tex]\begin{gathered} (h\cdot g)(x)=h(x)\cdot g(x) \\ (h\cdot g)(x)=(8x+512)(x^2-64) \end{gathered}[/tex]Apply the distributive property:
[tex]\begin{gathered} (h\cdot g)(x)=8x\cdot x^2+8x\cdot(-64)+512\cdot x^2+512\cdot(-64) \\ (h\cdot g)(x)=8x^3-512x+512x^2-32768 \\ And\text{ reorder terms:} \\ (h\cdot g)(x)=8x^3+512x^2-512x-32768 \end{gathered}[/tex]3. (g*h)(x):
[tex]\begin{gathered} (g\cdot h)(x)=g(x)\cdot h(x) \\ (g\cdot h)(x)=(x^2-64)\cdot\mleft(8x+512\mright) \end{gathered}[/tex]The result will be the same, but let's solve it to check. Apply the distributive property:
[tex]\begin{gathered} (g\cdot h)(x)=x^2\cdot8x+x^2\cdot512-64\cdot8x-64\cdot512 \\ (g\cdot h)(x)=8x^3+512x^2-512x-32768 \end{gathered}[/tex]