Question

find the absolute extrema of the function on the closed interval.
h(x)=2 - 3x^{2}, -2,1
minimum (x,y)=( )
maximum (x,y)=( )

Explanation:

Step1: Find the derivative of the function

The derivative of \(h(x)=2 - 3x^{2}\) is \(h^{\prime}(x)=-6x\).

Step2: Find the critical points

Set \(h^{\prime}(x) = 0\), so \(-6x=0\), which gives \(x = 0\).

Step3: Evaluate the function at the critical point and endpoints

• For \(x=-2\): \(h(-2)=2-3\times(-2)^{2}=2 - 12=-10\).
• For \(x = 0\): \(h(0)=2-3\times0^{2}=2\).
• For \(x = 1\): \(h(1)=2-3\times1^{2}=2 - 3=-1\).

Answer:

• minimum \((x,y)=(-2,-10)\)
• maximum \((x,y)=(0,2)\)