Question

Let f(x) = 9 - x, g (x) = x*2 + 2x - 8, and h (x) = x - 4

Answer 1

We are given the following functions

`\begin{gathered} f(x)=9-x \\ g(x)=x^2+2x-8 \\ h(x)=x-4 \end{gathered}`

g(x) + f(x)

`\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}`

h(x) - f(x)

`\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}`

f o h(10)

`\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}`

3 * g(-1)

`\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* g(-1) \\ \\ 3g(-1)=3*-9 \\ \\ 3g(-1)=-27 \end{gathered}`

h(x) * h(x)

`\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}`

g(x)/h(x)

`(g(x))/(h(x))=(x^2+2x-8)/(x-4),\text{ }x\\e4`