Problem 30 - 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. $$f(x)=12 x^{5}-20 x^{3} \text { on }[-2,2]$$

Step-by-Step Solution

Answer

Question: Determine the critical points of the function $$f(x) = 12x^5 - 20x^3$$ on the interval $$[-2, 2]$$, and indicate whether each critical point corresponds to a local maximum, local minimum, or neither. Answer: The critical points are $$x = -1$$ (local maximum), $$x = 0$$ (neither local maximum nor minimum), and $$x = 1$$ (local minimum).

Step 1: Find the first derivative of the function

To find the critical points, we need to find the first derivative of the function $$f(x)$$: $$f'(x) = \frac{d}{dx}(12x^5 - 20x^3)$$ Using the power rule, we find: $$f'(x) = 60x^4 - 60x^2$$

Step 2: Solve for the critical points

To find the critical points, we set the first derivative equal to zero and solve for $$x$$: $$60x^4 - 60x^2 = 0$$ Factor out a common factor of $$60x^2$$: $$60x^2(x^2 - 1) = 0$$ Now, we can find the three critical points by setting each factor equal to zero: $$x^2 = 0 \Rightarrow x = 0$$ $$x^2 - 1 = 0 \Rightarrow x = \pm 1$$ So the critical points are $$x = -1, x = 0$$ and $$x = 1$$.

Step 3: Analyze the function with a graphing utility

Now we will use a graphing utility (such as Desmos, GeoGebra, or any software) to graph the function $$f(x)$$ on the interval $$[-2, 2]$$ and visually determine the nature of each critical point. After graphing the function on the given interval, we can observe that: - At $$x=-1$$, there is a local maximum. - At $$x=0$$, it's neither a local maximum nor a local minimum, because it's an inflection point. - At $$x=1$$, there is a local minimum.

Step 4: Conclusion

The critical points of the function $$f(x)=12x^5-20x^3$$ on the interval $$[-2, 2]$$ are $$x=-1$$ (local maximum), $$x=0$$ (neither local maximum nor minimum), and $$x=1$$ (local minimum).

Key Concepts

First Derivative

The first derivative of a function, often denoted as \(f'(x)\), is a vital tool in calculus. It helps us determine the rate of change or the slope of a function at a given point. Calculating the first derivative is usually the first step in finding critical points, which tell us about the behavior of a function.

To find the first derivative, we apply rules such as the power rule, which involves bringing down the exponent as a coefficient and reducing the exponent by one. In the given exercise, the function \(f(x) = 12x^5 - 20x^3\) is differentiated using the power rule to form \(f'(x) = 60x^4 - 60x^2\).

• The term \(60x^4\) derives from \(12x^5\)
• The term \(60x^2\) derives from \(20x^3\)

Understanding this derivation is crucial for locating critical points, which involve setting this derivative equal to zero, indicating potential maxima, minima, or inflection points.

Local Maximum

A local maximum is a point on the graph of a function where the function changes direction from increasing to decreasing, making this point higher than surrounding points.

To identify a local maximum, one typically examines the sign changes in the first derivative. If \(f'(x) < 0\) going from left to right past the critical point, you might have found a local maximum. In simple terms, it's like reaching the top of a hill.

In the given function \(f(x) = 12x^5 - 20x^3\), the critical point at \(x = -1\) was identified as a local maximum. This was verified using a graphing utility or analyzing the signs of the first derivative around this point. The graph showed that the function changes from increasing to decreasing, confirming the hilltop scenario.

Local Minimum

A local minimum is the opposite of a local maximum. It's a point where the function changes direction from decreasing to increasing, making this point lower than the points around it.

To find a local minimum, examine the first derivative signs. If \(f'(x) > 0\) as you move past the critical point from left to right, it's a clue that you found a local minimum. Imagine this as reaching the bottom of a valley.

In the case of the function \(f(x) = 12x^5 - 20x^3\), the critical point at \(x = 1\) was identified as a local minimum. Graphical analysis or checking derivative signs confirmed that this point is lower compared to its neighbors, reflecting the valley scenario.

• First derivative changes sign from negative to positive at this point.

This understanding helps in visualizing how the function behaves around the critical points.

Inflection Point

An inflection point occurs where a function changes concavity, going from concave up to concave down or vice versa. At an inflection point, the first derivative might not change sign, which is why it's neither a local maximum nor a minimum.

Detecting an inflection point often involves looking at the second derivative, but in some problems, like our function \(f(x) = 12x^5 - 20x^3\), the first derivative can provide insights as well. At \(x = 0\), the graph flattens momentarily but does not form a peak or a valley.

In this exercise, the critical point at \(x = 0\) was not classified as a local maximum or minimum but precisely as an inflection point. Visual inspection of the graph emphasizes this behavior – the slope levels out, indicating a transition in curvature without an extremum. Understanding this nuance helps interpret transitions in function shapes effectively.