Question

Let g(x) = 3x, and h(x) = x2 + 1

Answer 1

Given:

`\begin{gathered} g(x)=3x \\ \\ h(x)=x^2+1 \end{gathered}`

Find:

`\begin{gathered} g(-1),\text{ }h(3),\text{ }g(g(-1)),\text{ }h(h(3)),\text{ }g(x)* h(x),\text{ }g(-2)* h(-2) \\ \end{gathered}`

Sol:

(a)

Value of g(-1)

`\begin{gathered} g(x)=3x \\ \\ g(-1)=3*(-1) \\ \\ g(-1)=-3 \\ \end{gathered}`

(b)

Value of h(3) is:

`\begin{gathered} h(x)=x^2+1 \\ \\ h(3)=3^2+1 \\ \\ h(3)=9+1 \\ \\ h(3)=10 \end{gathered}`

(c)

Value of g(g(-1))

`\begin{gathered} g(x)=3x \\ \\ g(g(x)) \\ \text{ } \\ \text{ Then }x=g(x) \\ \\ g(g(x)=3(3x) \\ \\ g(g(x))=9x \end{gathered}`

So the value of g(g(-1)) is:

`\begin{gathered} g(g(x))=9x \\ \\ g(g(-1))=9(-1) \\ \\ g(g(-1))=-9 \end{gathered}`

(c)

Value of h(h(3)) is:

`\begin{gathered} h(x)=x^2+1 \\ \\ x=h(x) \\ \\ \text{ Then,} \\ \\ h(h(x))=(x^2+1)^2+1 \\ \\ h(h(x))=x^4+2x^2+2 \\ \\ \end{gathered}`

So, the value of h(h(3)).

`\begin{gathered} h(h(x))=x^4+2x^2+2 \\ \\ h(h(3))=3^4+2(3)^2+2 \\ \\ =81+18+2 \\ \\ =101 \end{gathered}`

(d)

Value of g(x)*h(x)

`\begin{gathered} g(x)=3x \\ \\ h(x)=x^2+1 \\ \\ g(x)*h(x)=3x(x^2+1) \\ \\ g(x)^*h(x)=3x^3+3x \end{gathered}`

(e)

Value of g(-2)*h(-2)

`\begin{gathered} g(x)^*h(x)=3x^3+3x \\ \\ g(-2)^*h(-2)=3(-2)^3+3(-2) \\ \\ =-24-6 \\ \\ =-30 \end{gathered}`