We have the function:
[tex]h(x)=x^4-2x^2+4[/tex]
We can find the relative extrema by calculating its first derivative and equal to 0:
[tex]\begin{gathered} h^{\prime}(x)=4x^3-4x=0 \\ 4x(x^2-1)=0 \\ x^2-1=0 \\ x^2=1 \\ x=\pm1 \end{gathered}[/tex]
We have extremes at x=1 and x=-1. Also, at x=0 (when we clean 4x, it is because x=0 is a solution to the equation).
We can calculate the second derivative and evaluate each extreme to see if the extremes are maximums or minimums.
If the value of the second derivative is positive, it means that the extreme is a minimum. If it is negative, it is a maximum.
[tex]\begin{gathered} h^{\prime}^{\prime}(x)=12x^2-4 \\ h^{\prime\prime}(0)=12(0)^2-4=-4\longrightarrow\text{Maximum} \\ h^{\prime\prime}(-1)=12(-1)^2-4=12-4=8\longrightarrow\text{ Minimum} \\ h^{\prime\prime}(1)=12(1)^2-4=8\longrightarrow\text{ Minimum} \end{gathered}[/tex]
We will use this information to pick the correct graph:
- At x=0, the function has a value of h(0)=4.
- At x=0, the function has a local maximum.
- At x=-1 and x=1, the function has local minimums.
The correct graph is:
