Questions: Absolute max and min of f(x)=4x-x^2 on interval [1,4]

Solution Steps

To find the absolute maximum and minimum of the function \( f(x) = 4x - x^2 \) on the interval \([1, 4]\), we need to follow these steps:

1. Find the critical points by taking the derivative of \( f(x) \) and setting it to zero.
2. Evaluate the function at the critical points and at the endpoints of the interval.
3. Compare these values to determine the absolute maximum and minimum.

The function is given by \( f(x) = -x^2 + 4x \). To find the critical points, we first compute the derivative: \[ f'(x) = -2x + 4 \]

Setting the derivative equal to zero to find critical points: \[ -2x + 4 = 0 \implies 2x = 4 \implies x = 2 \] Thus, the critical point is \( x = 2 \).

Next, we evaluate \( f(x) \) at the critical point and the endpoints of the interval \([1, 4]\):

• At \( x = 1 \): \[ f(1) = -1^2 + 4 \cdot 1 = -1 + 4 = 3 \]
• At \( x = 2 \): \[ f(2) = -2^2 + 4 \cdot 2 = -4 + 8 = 4 \]
• At \( x = 4 \): \[ f(4) = -4^2 + 4 \cdot 4 = -16 + 16 = 0 \]

The values obtained are:

• \( f(1) = 3 \)
• \( f(2) = 4 \)
• \( f(4) = 0 \)

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

Final Answer