Problem 28 - Calculus Early Transcendentals

Question

a. Find the critical points of the following functions on the domain or on the given interval. b. Use a graphing utility to determine whether each critical point corresponds to a local maximum, local minimum, or neither. $$f(x)=\frac{4 x^{5}}{5}-3 x^{3}+5 \text { on }[-2,2]$$

Step-by-Step Solution

Answer

Question: Determine the nature of the critical points of the function $$f(x)=\frac{4 x^{5}}{5}-3 x^{3}+5$$ within the interval [-2, 2]. Answer: The critical points are \(x = -\frac{3}{\sqrt{20}}\), \(x=0\), and \(x = \frac{3}{\sqrt{20}}\). At \(x = -\frac{3}{\sqrt{20}}\) and \(x = \frac{3}{\sqrt{20}}\), the function has a local maximum, while at \(x = 0\), it has a local minimum.

Step 1: Find the derivative of the function

To find the critical points, we first need to find the derivative of the given function. The function is: $$f(x)=\frac{4 x^{5}}{5}-3 x^{3}+5$$ Differentiating the function with respect to x, we get: $$f'(x)=\frac{d}{dx}\left(\frac{4 x^{5}}{5}-3 x^{3}+5\right)$$ Applying the power rule and simplifying, we have: $$f'(x)=\frac{d}{dx}\left(4x^5\right)-\frac{d}{dx}\left(3x^3\right)=4(5x^4)-3(3x^2)=20x^4-9x^2$$

Step 2: Identify critical points

Now, we need to find the values of x where the derivative is either zero or undefined. As the derivative is a polynomial, it will always be defined. Therefore, we will focus on finding the values of x where the derivative is zero. So, solve for x in the equation: $$f'(x)=20x^4-9x^2=0$$ We observe that it's a quadratic equation in terms of \(x^2\). Let \(y = x^2\), the equation becomes: $$20y^2 - 9y = 0$$

Step 3: Solve for y

Now, solve the quadratic equation for y: $$20y^2 - 9y = y(20y - 9) = 0$$ So, the possible values for y are \(y = 0\) and \(y = \frac{9}{20}\).

Step 4: Solve for x

Now, as we have the possible values for y, find the corresponding values for x. 1. For \(y = 0\) (which means, \(x^2 = 0\)), we have \(x = 0\). 2. For \(y = \frac{9}{20}\) (which means, \(x^2 = \frac{9}{20}\)), we have \(x = \pm\sqrt{\frac{9}{20}} = \pm\frac{3}{\sqrt{20}}\). Since all these values lie within the given interval [-2, 2], they are valid critical points. So, the critical points are: $$x = -\frac{3}{\sqrt{20}}, 0, \frac{3}{\sqrt{20}}$$

Step 5: Graph the function and analyze critical points

Use a graphing utility to plot the function $$f(x)=\frac{4 x^{5}}{5}-3 x^{3}+5$$ on the interval \([-2,2]\). Observe the behavior of the function at the critical points found in Step 4. Looking at the graph, we can conclude the following: 1. At \(x = -\frac{3}{\sqrt{20}}\), the function has a local maximum. 2. At \(x = 0\), the function has a local minimum. 3. At \(x = \frac{3}{\sqrt{20}}\), the function has a local maximum. In conclusion, we have found the three critical points within the specified domain, and have determined their respective nature by using a graphing utility.

Key Concepts

Derivative

The derivative of a function is a fundamental concept in calculus used to determine the rate at which a function changes. In simpler terms, it shows how a small change in the input of a function (let's say an increase by a tiny amount) affects the output. When it comes to critical points, derivatives are key because they help us find where these points occur.

To find a derivative, you need to differentiate the function, which often involves rules and formulas. For polynomial functions, like the one in our exercise, the power rule is a handy tool to use. By applying the derivative, we can later determine points of maximum increase or decrease in the function.

Power Rule

The power rule is a simple yet powerful tool in calculus that aids in finding the derivative of polynomial expressions. This rule states that for any function of the form \( f(x) = ax^n \), the derivative is \( f'(x) = anx^{n-1} \).

For example, if we have a term \( 4x^5 \), applying the power rule gives us \( 20x^4 \). Each term in the function is approached this way: multiply the exponent with the coefficient, and then decrease the exponent by one.

• Makes finding derivatives of simple polynomials quick and efficient.
• Applies directly to each term in the polynomial separately, making it highly useful for longer expressions.

Using such rules allows us to systematically work through the polynomial to discover its rate of change which is crucial when analyzing function behavior for maxima and minima.

Local Maximum

A local maximum of a function is a point where the function value is higher than all other nearby values. In other words, it's like reaching the top of a hill when walking along a trail. In calculus terms, it's a critical point where the derivative changes from positive to negative.

For instance, at a local maximum, the function rises as x approaches the point and falls afterward. In our specific problem, we use the graph and derivative signs to identify around which critical point this happens. * Points are critical for understanding how functions behave in their domain. * They are identified using derivatives and often verified by graphing. Knowing these points gives insight into the function's overall increasing and decreasing behavior.

Local Minimum

Conversely, a local minimum is a point where the function value is lower than any other nearby values, much like finding the bottom of a valley. This is a critical point where the derivative changes from negative to positive.

If you're graphing a function and observe that it dips down into a trough, that dip represents a local minimum.

• These points show where the function switches from decreasing to increasing.
• Helps identify where the lowest output is locally compared to its surroundings.

In our given function, recognizing these points is crucial for understanding how the function behaves across different x-values, specifically understanding where the function climbs and descends sharply within the graphed interval.