let h(x)=x^4 - 2x^2. what is the absolute maximum value of h? choose 1 answer: a -1 b 0 c 1 d h has no…

Explanation:

Step1: Find the derivative

Differentiate $h(x)=x^{4}-2x^{2}$ using the power - rule. The derivative $h'(x)=4x^{3}-4x = 4x(x^{2}-1)=4x(x - 1)(x + 1)$.

Step2: Find critical points

Set $h'(x)=0$. Then $4x(x - 1)(x + 1)=0$. Solving this equation gives $x=-1,0,1$.

Step3: Evaluate the function at critical points

$h(-1)=(-1)^{4}-2(-1)^{2}=1 - 2=-1$. $h(0)=0^{4}-2\times0^{2}=0$. $h(1)=1^{4}-2\times1^{2}=1 - 2=-1$.

Step4: Analyze the behavior as $x\to\pm\infty$

As $x\to\pm\infty$, $y = h(x)=x^{4}-2x^{2}\to+\infty$ since the leading term $x^{4}$ has a positive coefficient and an even degree.

Answer:

D. $h$ has no maximum value