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

Question image

Understand the Problem

The question is asking to find the absolute maximum and minimum values of the function f(x) = x² within the interval [-2, 1]. This involves analyzing the function's critical points and evaluating its values at the endpoints of the interval.

Answer

The absolute maximum value is \( 4 \) at \( x = -2 \) and the absolute minimum value is \( 0 \) at \( x = 0 \).
Answer for screen readers

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

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 ): $$ f'(x) = 2x $$

  1. 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: $$ x = 0 $$

  1. 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 ): $$ f(1) = 1^2 = 1 $$
  1. Compare the function values

We now compare the values obtained:

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

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

More Information

The function ( f(x) = x^2 ) is a simple quadratic function that opens upwards. The critical point identified is a minimum since the graph of the function is U-shaped. The absolute maximum occurs at an endpoint in this case.

Tips

  • Neglecting to check endpoints: Some might only evaluate at the critical points, forgetting the endpoints, which can also hold the maximum or minimum values.
  • Not correctly calculating the derivative: Miscalculating the derivative can lead to incorrect critical points.

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