Consider The Function F ( X ) = 3 X 5 − 5 X 3 F(x) = 3x^5 - 5x^3 F ( X ) = 3 X 5 − 5 X 3 .3.1 Find The X X X -intercepts Of The Given Function F F F .3.2 Find The Critical Points.3.3 Find The Regions Of Increase And Decrease For The Given Function F F F .

Analyzing the Function $f(x) = 3x^5 - 5x^3$

3.1 Finding the $x$-intercepts of the given function $f$

To find the $x$-intercepts of the function $f(x) = 3x^5 - 5x^3$, we need to set the function equal to zero and solve for $x$. This is because the $x$-intercepts are the points where the graph of the function crosses the $x$-axis, and at these points, the value of the function is zero.

Setting the function equal to zero, we get:

$$3x^5 - 5x^3 = 0$$

We can factor out an $x^3$ from both terms:

$$x^3(3x^2 - 5) = 0$$

This tells us that either $x^3 = 0$ or $3x^2 - 5 = 0$. Solving for $x$ in both cases, we get:

$$x^3 = 0 \Rightarrow x = 0$$

$$3x^2 - 5 = 0 \Rightarrow x^2 = \frac{5}{3} \Rightarrow x = \pm \sqrt{\frac{5}{3}}$$

Therefore, the $x$-intercepts of the function $f(x) = 3x^5 - 5x^3$ are $x = 0$, $x = \sqrt{\frac{5}{3}}$, and $x = -\sqrt{\frac{5}{3}}$.

3.2 Finding the critical points

To find the critical points of the function $f(x) = 3x^5 - 5x^3$, we need to find the values of $x$ where the derivative of the function is equal to zero or undefined.

First, let's find the derivative of the function using the power rule:

$$f'(x) = 15x^4 - 15x^2$$

Now, we need to set the derivative equal to zero and solve for $x$:

$$15x^4 - 15x^2 = 0$$

We can factor out a $15x^2$ from both terms:

$$15x^2(x^2 - 1) = 0$$

This tells us that either $15x^2 = 0$ or $x^2 - 1 = 0$. Solving for $x$ in both cases, we get:

$$15x^2 = 0 \Rightarrow x = 0$$

$$x^2 - 1 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$$

Therefore, the critical points of the function $f(x) = 3x^5 - 5x^3$ are $x = 0$, $x = 1$, and $x = -1$.

3.3 Finding the regions of increase and decrease

To find the regions of increase and decrease for the function $f(x) = 3x^5 - 5x^3$, we need to examine the sign of the derivative of the function.

Recall that the derivative of the function is:

$$f'(x) = 15x^4 - 15x^2$$

We can factor out a $15x^2$ from both terms:

$$f'(x) = 15x^2(x^2 - 1)$$

Now, we can examine the sign of the derivative in different intervals.

For $x < -1$, both $x^2$ and $(x^2 - 1)$ are negative, so the derivative is positive.

For $-1 < x < 0$, $x^2$ is positive, but $(x^2 - 1)$ is negative, so the derivative is negative.

For $0 < x < 1$, both $x^2$ and $(x^2 - 1)$ are positive, so the derivative is positive.

For $x > 1$, both $x^2$ and $(x^2 - 1)$ are positive, so the derivative is positive.

Therefore, the function $f(x) = 3x^5 - 5x^3$ is increasing on the intervals $(-\infty, -1)$, $(0, 1)$, and $(1, \infty)$, and decreasing on the interval $(-1, 0)$.

Conclusion

In this article, we analyzed the function $f(x) = 3x^5 - 5x^3$ and found its $x$-intercepts, critical points, and regions of increase and decrease. We used the power rule to find the derivative of the function and examined the sign of the derivative in different intervals to determine the regions of increase and decrease.

Key Takeaways

• The $x$-intercepts of the function $f(x) = 3x^5 - 5x^3$ are $x = 0$, $x = \sqrt{\frac{5}{3}}$, and $x = -\sqrt{\frac{5}{3}}$.
• The critical points of the function $f(x) = 3x^5 - 5x^3$ are $x = 0$, $x = 1$, and $x = -1$.
• The function $f(x) = 3x^5 - 5x^3$ is increasing on the intervals $(-\infty, -1)$, $(0, 1)$, and $(1, \infty)$, and decreasing on the interval $(-1, 0)$.

References

• [1] Calculus, 3rd edition, Michael Spivak
• [2] Calculus, 2nd edition, James Stewart
  Q&A: Analyzing the Function $f(x) = 3x^5 - 5x^3$

Q: What are the $x$-intercepts of the function $f(x) = 3x^5 - 5x^3$?

A: The $x$-intercepts of the function $f(x) = 3x^5 - 5x^3$ are $x = 0$, $x = \sqrt{\frac{5}{3}}$, and $x = -\sqrt{\frac{5}{3}}$. These are the points where the graph of the function crosses the $x$-axis, and at these points, the value of the function is zero.

Q: How do you find the critical points of the function $f(x) = 3x^5 - 5x^3$?

A: To find the critical points of the function $f(x) = 3x^5 - 5x^3$, we need to find the values of $x$ where the derivative of the function is equal to zero or undefined. We can find the derivative of the function using the power rule, and then set the derivative equal to zero and solve for $x$.

Q: What are the critical points of the function $f(x) = 3x^5 - 5x^3$?

A: The critical points of the function $f(x) = 3x^5 - 5x^3$ are $x = 0$, $x = 1$, and $x = -1$. These are the points where the derivative of the function is equal to zero or undefined.

Q: How do you determine the regions of increase and decrease for the function $f(x) = 3x^5 - 5x^3$?

A: To determine the regions of increase and decrease for the function $f(x) = 3x^5 - 5x^3$, we need to examine the sign of the derivative of the function. We can factor out a $15x^2$ from both terms of the derivative, and then examine the sign of the derivative in different intervals.

Q: What are the regions of increase and decrease for the function $f(x) = 3x^5 - 5x^3$?

A: The function $f(x) = 3x^5 - 5x^3$ is increasing on the intervals $(-\infty, -1)$, $(0, 1)$, and $(1, \infty)$, and decreasing on the interval $(-1, 0)$.

Q: What is the significance of the $x$-intercepts, critical points, and regions of increase and decrease for the function $f(x) = 3x^5 - 5x^3$?

A: The $x$-intercepts, critical points, and regions of increase and decrease for the function $f(x) = 3x^5 - 5x^3$ are important because they help us understand the behavior of the function. The $x$-intercepts tell us where the graph of the function crosses the $x$-axis, the critical points tell us where the derivative of the function is equal to zero or undefined, and the regions of increase and decrease tell us where the function is increasing or decreasing.

Q: How can you use the information about the $x$-intercepts, critical points, and regions of increase and decrease to graph the function $f(x) = 3x^5 - 5x^3$?

A: To graph the function $f(x) = 3x^5 - 5x^3$, you can use the information about the $x$-intercepts, critical points, and regions of increase and decrease to determine the shape of the graph. You can start by plotting the $x$-intercepts and critical points, and then use the information about the regions of increase and decrease to determine the direction of the graph.

Q: What are some real-world applications of the function $f(x) = 3x^5 - 5x^3$?

A: The function $f(x) = 3x^5 - 5x^3$ has many real-world applications, including modeling population growth, chemical reactions, and electrical circuits. It can also be used to model the behavior of complex systems, such as financial markets and social networks.

Q: How can you use the function $f(x) = 3x^5 - 5x^3$ to model real-world phenomena?

A: To use the function $f(x) = 3x^5 - 5x^3$ to model real-world phenomena, you can substitute real-world data into the function and use it to make predictions about the behavior of the system. For example, you can use the function to model the growth of a population over time, or the behavior of a chemical reaction.

Conclusion

In this Q&A article, we discussed the $x$-intercepts, critical points, and regions of increase and decrease for the function $f(x) = 3x^5 - 5x^3$. We also discussed some real-world applications of the function and how it can be used to model complex systems.