Solution
We are given the following functions
[tex]\begin{gathered} f(x)=9-x \\ g(x)=x^2+2x-8 \\ h(x)=x-4 \end{gathered}[/tex]
g(x) + f(x)
[tex]\begin{gathered} g(x)+f(x)=(x^2+2x-8)+(9-x) \\ \\ g(x)+f(x)=x^2+2x-8+9-x \\ \\ g(x)+f(x)=x^2+x+1 \end{gathered}[/tex]
h(x) - f(x)
[tex]\begin{gathered} h(x)-f(x)=(x-4)-(9-x) \\ \\ h(x)-f(x)=x-4-9+x \\ \\ h(x)-f(x)=2x-13 \end{gathered}[/tex]
f o h(10)
[tex]\begin{gathered} First \\ h(x)=x-4 \\ h(10)=10-4 \\ h(10)=6 \\ and \\ f(x)=9-x \\ f(6)=9-6 \\ f(6)=3 \\ Now,\text{ to solve} \\ foh(10)=f(h(10)) \\ foh(10)=f(6) \\ \\ foh(10)=3 \end{gathered}[/tex]
3 * g(-1)
[tex]\begin{gathered} First, \\ g(x)=x^2+2x-8 \\ g(-1)=(-1)^2+2(-1)-8 \\ \\ g(-1)=1-2-8 \\ \\ g(-1)=-9 \\ Now\text{ to solve} \\ 3g(-1)=3\times g(-1) \\ \\ 3g(-1)=3\times-9 \\ \\ 3g(-1)=-27 \end{gathered}[/tex]
h(x) * h(x)
[tex]\begin{gathered} h(x)=x-4 \\ Now, \\ h(x)*h(x)=(x-4)(x-4) \\ \\ h(x)*h(x)=x^2-8x+16 \end{gathered}[/tex]
g(x)/h(x)
[tex]\frac{g(x)}{h(x)}=\frac{x^2+2x-8}{x-4},\text{ }x\ne4[/tex]
