use a graphing calculator to graph each function and to determine and classify its extrema. 22…

Explanation:

Step1: Find the derivative of $h(x)$

Differentiate $h(x)=x^{4}-4x^{2}-2$ using power - rule. $h'(x)=4x^{3}-8x = 4x(x^{2}-2)=4x(x - \sqrt{2})(x+\sqrt{2})$.

Step2: Find the critical points

Set $h'(x)=0$. Then $4x(x - \sqrt{2})(x+\sqrt{2})=0$. The critical points are $x = 0,x=\sqrt{2},x=-\sqrt{2}$.

Step3: Find the second - derivative of $h(x)$

Differentiate $h'(x)=4x^{3}-8x$ to get $h''(x)=12x^{2}-8$.

Step4: Classify the critical points of $h(x)$

• For $x = 0$: $h''(0)=12\times0^{2}-8=-8<0$, so $h(x)$ has a local maximum at $x = 0$ and $h(0)=0^{4}-4\times0^{2}-2=-2$.
• For $x=\sqrt{2}$: $h''(\sqrt{2})=12\times(\sqrt{2})^{2}-8=12\times2 - 8 = 16>0$, so $h(x)$ has a local minimum at $x=\sqrt{2}$ and $h(\sqrt{2})=(\sqrt{2})^{4}-4\times(\sqrt{2})^{2}-2=4 - 8 - 2=-6$.
• For $x=-\sqrt{2}$: $h''(-\sqrt{2})=12\times(-\sqrt{2})^{2}-8=16>0$, so $h(x)$ has a local minimum at $x=-\sqrt{2}$ and $h(-\sqrt{2})=(-\sqrt{2})^{4}-4\times(-\sqrt{2})^{2}-2=-6$.

For $y=-x^{2}+8x - 10$:

Step1: Find the first - derivative

$y'=-2x + 8$.

Step2: Find the second - derivative

$y''=-2$.

Step3: Classify the critical point $x = 4$

Since $y''=-2<0$ when $x = 4$, the function $y=-x^{2}+8x - 10$ has a local maximum at $x = 4$. And $y(4)=-4^{2}+8\times4 - 10=-16 + 32-10 = 6$.

Answer:

For $h(x)=x^{4}-4x^{2}-2$, local maximum at $(0,-2)$, local minima at $(\pm\sqrt{2},-6)$. For $y=-x^{2}+8x - 10$, local maximum at $(4,6)$.