#### Answer

The maximum value of the function is $7$ is at attained when $x=\pm\sqrt{2}$.

#### Work Step by Step

As suggested, we would use the substitution $t = x^2$ to get $f(t) = 3+4t-t^2$. To find the maximum of this quadratic function, we find the vertex using the formulas derived in the chapter, mainly $h=\frac{-b}{2a}$ and $k = f(h)$ to get $h=2$ and $k = f(2) = 7.$
We note that this would be the solution for the new equation $f(t)$ rather than $f(x)$.
To get the solution for $f(x)$, we plug back in $t=x^2$ to get $x=\pm\sqrt{t}$ hence the maximum value of the function is $7$ and is at attained at $x=\pm\sqrt{2}$.
Attached is the graph of both $f(x)$ and $f(t)$ to illustrate the results with the points above marked as well.