Questions: Find the absolute maximum and absolute minimum values of f(x)=2x^(1 / 2) on the interval [-1,1].

Solution Steps

To find the absolute maximum and minimum values of the function \( f(x) = 2|x|^{1/2} \) on the interval \([-1, 1]\), we need to evaluate the function at critical points and endpoints. The critical points occur where the derivative is zero or undefined. Since the function involves an absolute value, we consider the behavior on the intervals \([-1, 0]\) and \([0, 1]\). We then compare the function values at these points to determine the absolute extrema.

We are given the function \( f(x) = 2|x|^{1/2} \) and need to find its absolute maximum and minimum values on the interval \([-1, 1]\).

First, we evaluate the function at the endpoints of the interval:

• At \( x = -1 \): \[ f(-1) = 2|-1|^{1/2} = 2 \times 1 = 2 \]

• At \( x = 1 \): \[ f(1) = 2|1|^{1/2} = 2 \times 1 = 2 \]

Next, we find the critical points by considering the derivative of \( f(x) \). Since \( f(x) = 2|x|^{1/2} \), we need to consider the derivative separately for \( x \geq 0 \) and \( x < 0 \).

For \( x \geq 0 \), \( f(x) = 2x^{1/2} \): \[ f'(x) = 2 \cdot \frac{1}{2}x^{-1/2} = x^{-1/2} = \frac{1}{\sqrt{x}} \]

For \( x < 0 \), \( f(x) = 2(-x)^{1/2} \): \[ f'(x) = 2 \cdot \frac{1}{2}(-x)^{-1/2} \cdot (-1) = -(-x)^{-1/2} = -\frac{1}{\sqrt{-x}} \]

The derivative does not exist at \( x = 0 \) because the function is not differentiable at that point due to the absolute value.

Since the derivative does not exist at \( x = 0 \), we evaluate the function at this point:

• At \( x = 0 \): \[ f(0) = 2|0|^{1/2} = 0 \]

From the evaluations, we have:

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

The absolute maximum value is 2, occurring at \( x = -1 \) and \( x = 1 \).

The absolute minimum value is 0, occurring at \( x = 0 \).

Final Answer