Question

a. Locate the critical points of \(f\) b. Use the First Derivative Test to locate the local maximum and minimum values. c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist). $$f(x)=x \sqrt{4-x^{2}} \text { on }[-2,2]$$

Short answer

Answer: The absolute maximum value of the function on the interval \([-2, 2]\) is \(\sqrt{3}\), and the absolute minimum value is \(0\).

Step-by-step solution

Find the first derivative of the function f(x)

First, we have to find the derivative of the function, which will be used for the First Derivative Test and for finding the critical points. We have to use the product rule for differentiation, which states that for \(u(x) \cdot v(x)\), the derivative is given by: $$(u \cdot v)'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$$ Let \(u(x) = x\) and \(v(x) = \sqrt{4 - x^2}\). Then: $$u'(x) = 1$$ $$v'(x) = \frac{-2x}{2\sqrt{4 - x^2}} = \frac{-x}{\sqrt{4 - x^2}}$$ Now, applying the product rule: $$f'(x) = (1) \cdot (\sqrt{4 - x^2}) + (x) \cdot \left(\frac{-x}{\sqrt{4 - x^2}}\right)$$ $$f'(x) = \sqrt{4 - x^2} - \frac{x^2}{\sqrt{4 - x^2}}$$

Find critical points

Critical points are points where the first derivative is either equal to zero or undefined. To find the critical points, set \(f'(x)\) equal to zero and solve for \(x\): $$0 = \sqrt{4 - x^2} - \frac{x^2}{\sqrt{4 - x^2}}$$ To solve for x, we'll first get a common denominator: $$0 = \frac{4 - x^2 - x^2(4 - x^2)}{4 - x^2}$$ Simplifying the numerator, we get: $$0 = \frac{4 - x^2 - 4x^2 + x^4}{4 - x^2}$$ $$0 = \frac{x^4 - 5x^2 + 4}{4 - x^2}$$ Now, we notice that the numerator is a quadratic in \(x^2\), so we can let \(y = x^2\) and solve for \(y\): $$0 = y^2 - 5y + 4$$ This is a quadratic equation which we can factor: $$0 = (y - 4)(y - 1)$$ Solving for \(y\) (which is \(x^2\)), we get \(x^2 = 1\) and \(x^2 = 4\), so \(x = \pm 1\) and \(x = \pm 2\). The critical points are at $$x = -2, -1, 1, 2$$

Use the First Derivative Test

Now, we'll use the first derivative test to determine if the critical points correspond to local maximum or minimum values. Test the intervals around the critical points to see if the function is increasing or decreasing: - Interval \((-\infty, -2)\): Choose \(x = -3\). Then, \(f'(-3)\) is undefined because it is outside the domain of the function. We cannot conclude anything from this interval. - Interval \((-2, -1)\): Choose \(x = -\frac{3}{2}\). Then, \(f'(-\frac{3}{2}) > 0\), so the function is increasing in this interval. - Interval \((-1,1)\): Choose \(x = 0\). Then, \(f'(0) < 0\), so the function is decreasing in this interval. - Interval \((1,2)\): Choose \(x = \frac{3}{2}\). Then, \(f'(\frac{3}{2}) > 0\), so the function is increasing in this interval. - Interval \((2, \infty)\): same situation as \((-\infty, -2)\). From the first derivative test: - \(x = -2\): local minimum - \(x = -1\): local maximum - \(x = 1\): local maximum - \(x = 2\): local minimum

Identify the absolute maximum and minimum values

Now, we will evaluate the function at the critical points and at the endpoints of the interval to find the absolute maximum and minimum values. We already have \(f(-2) = 0\), \(f(-1) = \sqrt{3}\), \(f(1) = \sqrt{3}\), and \(f(2) = 0\). Thus, the absolute maximum value of the function on the interval \([-2, 2]\) is $$\sqrt{3}$$, and the absolute minimum value is $$0$$.

Key concepts

Critical Points

In calculus, critical points are where a function's derivative is zero or undefined. These points are essential in finding where a function might be at a local maximum or minimum. For the function \(f(x) = x \sqrt{4 - x^2}\), we used the product rule to find the derivative:

• First, determine the derivative \(f'(x)\) using the product and chain rules.
• Next, set \(f'(x) = 0\) or find where it's undefined to locate the critical points.

For our example, the critical points were found at \(x = -2, -1, 1, 2\). These are the x-values where changes in the function's increasing or decreasing behavior might occur.

First Derivative Test

The First Derivative Test helps us determine if critical points are local maxima or minima. By evaluating the sign of the derivative around these points:

• If the derivative changes from positive to negative, you have a local maximum.
• If it changes from negative to positive, you have a local minimum.
• No change usually means it's neither a maximum nor a minimum.

In the example, by testing intervals around critical points, we discovered:

• \(x = -2\) and \(x = 2\) are local minima.
• \(x = -1\) and \(x = 1\) are local maxima.

This test provides insight into the behavior of the function at various points.

Absolute Maximum and Minimum

Absolute maximum and minimum values are the highest and lowest points over an entire interval.To find these for \(f(x)\) on \([-2, 2]\):

• Evaluate \(f(x)\) at the critical points and the endpoints of the interval.
• Compare these values to determine the overall highest and lowest values.

For \(f(x) = x \sqrt{4 - x^2}\), the critical points gave us \(f(-1) = \sqrt{3}\) and \(f(1) = \sqrt{3}\), while the endpoints \(f(-2)\) and \(f(2)\) evaluated to \(0\).Thus, on the interval \([-2,2]\), the absolute maximum value is \(\sqrt{3}\), while the absolute minimum is \(0\). Identifying these values is crucial for understanding the function's behavior across the entire range.