Answer
$-14$
Work Step by Step
Given: $f(x)=2+16x^3+4x^6$
Let us consider $ t=x^3, t^2=x^6$
Now, we have $f(t)=4t^2+16t+2$
Here, $a=4, b=16, c=2$
But $a=4 \gt 0$ shows that the function has a minimu value.
The maximum value occurs at $t=-\dfrac{b}{2a}=-\dfrac{16}{2(4)}=-2$
Thus, the maximum value of the function $f(x)$ at $t=-2$ is:
$f(-2)=4(-2)^2+16(-2)+2$
This gives:
$f(-2)=-14$
