Find the absolute maximum and minimum values of f(x) = x² on [-2, 1].
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
- Find the derivative of the function
To determine the critical points, we first take the derivative of the function ( f(x) = x^2 ): $$ f'(x) = 2x $$
- 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: $$ x = 0 $$
- 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 ): $$ f(1) = 1^2 = 1 $$
- Compare the function values
We now compare the values obtained:
- ( f(-2) = 4 )
- ( f(0) = 0 )
- ( f(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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