Solve number 2

Step 1: Complete part e) of question 2. Given $g(x) = \frac{x}{3x+2}$ and $h(x) = \sqrt{x^2 - 1}$.

e) Find $g(a^2) h(a^2)$. First, find $g(a^2)$: $g(a^2) = \frac{a^2}{3(a^2)+2} = \frac{a^2}{3a^2+2}$ Next, find $h(a^2)$: $h(a^2) = \sqrt{(a^2)^2 - 1} = \sqrt{a^4 - 1}$ Now, multiply $g(a^2)$ and $h(a^2)$: $g(a^2) h(a^2) = \frac{a^2}{3a^2+2} \cdot \sqrt{a^4 - 1}$ The value is $\boxed{\frac{a^2\sqrt{a^4 - 1}}{3a^2+2}}$.