find the absolute extrema of the function on the closed interval.\nh(x)=2 - 3x^{2}, -2,1\nminimum (x,y)=(…

find the absolute extrema of the function on the closed interval.\nh(x)=2 - 3x^{2}, -2,1\nminimum (x,y)=( )\nmaximum (x,y)=( )

find the absolute extrema of the function on the closed interval.\nh(x)=2 - 3x^{2}, -2,1\nminimum (x,y)=( )\nmaximum (x,y)=( )

Answer

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))