Find the absolute maximum and minimum values of f(x) = x² on [-2, 1].

Steps to Solve

1. Find the derivative of the function

To determine the critical points, we first take the derivative of the function ( f(x) = x^2 ): $$ f'(x) = 2x $$

2. Set the derivative equal to zero

Next, we find the values of ( x ) where the derivative is zero, which gives us the critical points: $$ 2x = 0 $$ This leads to: $$ x = 0 $$

3. Evaluate the function at the critical points and endpoints

Now we evaluate ( f(x) ) at the critical point ( x = 0 ) and at the endpoints of the interval ( x = -2 ) and ( x = 1 ):

• At ( x = -2 ): $$ f(-2) = (-2)^2 = 4 $$
• At ( x = 0 ): $$ f(0) = 0^2 = 0 $$
• At ( x = 1 ): $$ f(1) = 1^2 = 1 $$

4. Compare the function values

We now compare the values obtained:

• ( f(-2) = 4 )
• ( f(0) = 0 )
• ( f(1) = 1 )

5. Identify the absolute maximum and minimum

From the evaluated values, we determine:

• The absolute maximum value is ( 4 ) at ( x = -2 ).
• The absolute minimum value is ( 0 ) at ( x = 0 ).