Answer
$7$
Work Step by Step
Given: $f(x)=3+4x^2-x^4$
Let us consider $x=t, t=x^2, t^2=x^4$
Now, we have $f(t)=-t^2+4t+3$
Here, $a=-1, b=4, c=3$
But $a=-1 \lt 0$ shows that the function has a maximum value.
The maximum value occurs at $t=-\dfrac{b}{2a}=-\dfrac{4}{2(-1)}=2$
Thus, the maximum value of the function $f(x)=3+4x^2-x^4$ occurs is $f(-\dfrac{b}{2a})$.
$f(2)=3+4(2)-(2)^2$
This gives:
$f(2)=7$