Chapter 3.1, Problem 80E

To determine

To find: The maximum value of the function $f\left(x\right)=3+4x^2-x^4$ .

Answer to Problem 80E

The maximum value of the function $f\left(x\right)=3+4x^2-x^4$ is equal to $7$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=3+4x^2-x^4$ .

Calculation:

Substitute $t$ for $x^2$ in the given function.

  $\begin{array}{c}f\left(x\right)=3+4x^2-x^4\\ f\left(x\right)=3+4x^2-{\left(x^2\right)}^2\\ f\left(t\right)=3+4t-t^2\end{array}$

Compare the quadratic function in $t$ with $a{t}^2+bt+c$ . This gives $a=-1$ , $b=4$ and $c=3$ .

The maximum value of the function in $t$ occurs at $t=-\frac{b}{2a}$ i.e. $t=-\frac{4}{-2}=2$ .

So, the maximum value of the function $f\left(t\right)=3+4t-t^2$ occurs at $t=2$ .

Substitute $2$ for $t$ in the function.

  $\begin{array}{c}f\left(t\right)=3+4t-t^2\\ f\left(2\right)=3+4\left(2\right)-{\left(2\right)}^2\\ f\left(2\right)=3+8-4\\ f\left(2\right)=7\end{array}$

The maximum value of the function $f\left(t\right)=3+4t-t^2$ is equal to $7$ and also the maximum value of the function $f\left(x\right)=3+4x^2-x^4$ is equal to $7$ at $x=\sqrt{2}=\pm 1.414$ .

Therefore, the maximum value of the function $f\left(x\right)=3+4x^2-x^4$ is equal to $7$ .