Question

a. Find the critical points of \(f\) on the given interval. b. Determine the absolute extreme values of \(f\) on the given interval when they exist. c. Use a graphing utility to confirm your conclusions. $$f(x)=2 x^{6}-15 x^{4}+24 x^{2} \text { on }[-2,2]$$

Short answer

Question: Determine the absolute maximum and minimum values of the function \(f(x) = 2x^6 - 15x^4 + 24x^2\) on the interval \([-2, 2]\). Answer: The absolute maximum value is 28, occurring at x=-2 and x=2. The absolute minimum value is 0, occurring at x=0.

Step-by-step solution

Compute the derivative of the function

Find the derivative of the function \(f(x)=2 x^{6}-15 x^{4}+24 x^{2}\) with respect to x. To do that, we will apply the power rule for derivatives \((d/dx(x^n)=nx^{n-1})\). We get: $$f'(x)=12x^5-60x^3+48x$$

Find the critical points of the function

To find the critical points, we need to set the derivative equal to zero and solve for x: $$12x^5-60x^3+48x=0$$ Factor out the common term x to simplify the equation: $$x(12x^4-60x^2+48)=0$$ We have the first critical point: \(x_1=0\) Now, we need to solve the quadratic equation \(12x^4-60x^2+48=0\). Divide the equation by 12 to simplify: $$x^4-5x^2+4=0$$ We will use the substitution \(u = x^2\), which transforms the equation into a quadratic equation: $$u^2-5u+4=0$$ Factor the quadratic equation: $$(u-4)(u-1)=0$$ The solutions are \(u_1=1\) and \(u_2=4\). Substitute back x^2 for u to find the other critical points: $$x^2=1 \Rightarrow x_2 =-1, x_3=1$$ $$x^2=4 \Rightarrow x_4 =-2, x_5=2$$

Determine absolute extreme values of the function

To find the absolute extreme values, evaluate the function at its critical points as well as the endpoints of the interval. The critical points are \(x_1=0\), \(x_2=-1\), \(x_3=1\), \(x_4=-2\), and \(x_5=2\). The interval is \([-2, 2]\). By plugging in these values into the original function, we get: $$f(-2)=64-60+24=28$$ $$f(-1)=2-15+24=11$$ $$f(0)=0$$ $$f(1)=2-15+24=11$$ $$f(2)=64-60+24=28$$

Identify the absolute extreme values

After evaluating the function at the critical points and endpoints, we can compare the function values to determine the absolute maximum and minimum: Absolute maximum is 28, occurring at x=-2 and x=2. Absolute minimum is 0, occurring at x=0.

Confirm the conclusions using a graphing utility

Using a graphing utility, plot \(f(x)=2 x^{6}-15 x^{4}+24 x^{2}\) on the interval \([-2, 2]\). You will see that the function has its lowest point at x=0 and highest points at x=-2 and x=2, confirming our conclusions.

Key concepts

Understanding Derivatives

To grasp the idea of critical points and extreme values, it's essential to understand derivatives first. A derivative represents the rate at which a function changes. It provides a way to capture how a function's output value alters as the input changes slightly. When we take the derivative of a function, we're essentially computing its slope at various points.

In this exercise, you're given a polynomial function, and you need to find its derivative. For functions like this, you use the power rule of derivatives. This rule states that for any term of the form \( x^n \), the derivative is \( nx^{n-1} \). Applying this rule helps you find where the function's slope is zero, which are potential critical points for extreme values.

In our case, the derivative was calculated as \( f'(x) = 12x^5 - 60x^3 + 48x \). This equation tells us how the function behaves at different points along the x-axis. By finding when \( f'(x) \, = \, 0 \), we locate the function's critical points, which are key to identifying extreme values.

Determining Extreme Values

Now with the derivative in hand, you can find where the function reaches its extreme values on an interval. Our method involves evaluating the original function at critical points where the derivative equals zero, and also at the endpoints of the given interval.

Here's a step-by-step breakdown:

• First, solve the derivative equation \( f'(x) = 0 \) to find the x-values of the critical points.
• Next, substitute these x-values back into the original function to determine their corresponding y-values, which are potential extreme values.
• Finally, compare these values with the function values at the endpoints of the interval to identify the absolute maximum and minimum values.

To illustrate, in the step-by-step solution, the critical points found were \( x = -2, -1, 0, 1, 2 \). By evaluating the original function at these points, it was determined that the absolute maximum value is 28 and occurs at \( x = -2 \) and \( x = 2 \). The absolute minimum value is 0 at \( x = 0 \). This practical approach allows you to systematically find where a function takes its peak and valley values within a specific range.

Exploring Polynomial Functions

Polynomial functions like \( f(x) = 2x^6 - 15x^4 + 24x^2 \) are combinations of terms called monomials, each consisting of a variable raised to a whole number power. These functions are particularly interesting because they are smooth, continuous, and differentiable, which means they can provide good opportunities for analyzing changes and trends depicted by their curves.

The degree of a polynomial (the highest power of the variable present) majorly influences its behavior. For instance, even-degree polynomials, like our example's 6th degree, often have certain symmetrical properties and typically end pointing upwards or downwards at both of the axis limits of the graph. This can create multiple highs and lows within the function's graph.

• A 6th-degree polynomial may have up to five critical points, representing changes in direction.
• These points result from setting the derivative of the function to zero and solving, just like in our example.

Graphing these functions can reveal the number of peaks and valleys, helping us visualize where these slopes change and denote extreme points we calculated. Overall, studying polynomial functions requires an understanding of their structure, degree, and resulting behavior upon graphing or analyzing.