Part 3 Of 4Answer The Following Questions About The Function Whose Derivative Is Given Below.a. What Are The Critical Points Of F F F ?b. On What Open Intervals Is F F F Increasing Or Decreasing?c. At What Points, If Any, Does F F F

Understanding the Function and Its Derivative

To answer the questions about the function whose derivative is given, we first need to understand the relationship between the function and its derivative. The derivative of a function represents the rate of change of the function with respect to its input variable. In this case, we are given the derivative of a function $f$, and we need to use this information to determine the critical points of $f$, the intervals on which $f$ is increasing or decreasing, and the points at which $f$ has a local maximum or minimum.

The Derivative of the Function

The derivative of the function $f$ is given by:

$$f'(x) = 2x^2 - 6x + 9$$

Finding the Critical Points

To find the critical points of $f$, we need to find the values of $x$ for which the derivative $f'(x)$ is equal to zero or undefined. In this case, the derivative is a polynomial function, so it is defined for all real numbers. Therefore, we need to find the values of $x$ for which $f'(x) = 0$.

To do this, we can set the derivative equal to zero and solve for $x$:

$$2x^2 - 6x + 9 = 0$$

We can solve this quadratic equation using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 2$, $b = -6$, and $c = 9$. Plugging these values into the quadratic formula, we get:

$$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(9)}}{2(2)}$$

Simplifying this expression, we get:

$$x = \frac{6 \pm \sqrt{36 - 72}}{4}$$

$$x = \frac{6 \pm \sqrt{-36}}{4}$$

Since the square root of a negative number is not a real number, this equation has no real solutions. Therefore, the derivative $f'(x)$ is never equal to zero, and the function $f$ has no critical points.

Intervals of Increase and Decrease

Since the derivative $f'(x)$ is never equal to zero, we can conclude that the function $f$ is either always increasing or always decreasing. To determine which of these is the case, we can examine the sign of the derivative.

The derivative $f'(x) = 2x^2 - 6x + 9$ is a quadratic function, and its graph is a parabola that opens upward. Since the coefficient of the $x^2$ term is positive, the parabola is concave upward, and the derivative is positive for all values of $x$.

Therefore, we can conclude that the function $f$ is always increasing.

Local Maxima and Minima

Since the function $f$ is always increasing, it has no local maxima or minima. A local maximum or minimum occurs when the function changes from increasing to decreasing or from decreasing to increasing, respectively. However, since the function $f$ is always increasing, it never changes from increasing to decreasing, and therefore it has no local maxima or minima.

Conclusion

In conclusion, the function $f$ has no critical points, and it is always increasing. Therefore, it has no local maxima or minima.

Key Takeaways

• The critical points of a function are the values of the input variable at which the derivative is equal to zero or undefined.
• The intervals on which a function is increasing or decreasing can be determined by examining the sign of the derivative.
• A function has a local maximum or minimum at a point where the function changes from increasing to decreasing or from decreasing to increasing, respectively.

Further Reading

For further reading on this topic, we recommend the following resources:

• Calculus: Early Transcendentals
• Calculus: Single Variable
• Mathematics for Calculus

Practice Problems

To practice solving problems like this, we recommend the following exercises:

• Find the critical points of the function $f(x) = x^2 - 4x + 3$.
• Determine the intervals on which the function $f(x) = x^2 - 4x + 3$ is increasing or decreasing.
• Find the local maxima and minima of the function $f(x) = x^2 - 4x + 3$.
  

Understanding the Function and Its Derivative

To answer the questions about the function whose derivative is given, we first need to understand the relationship between the function and its derivative. The derivative of a function represents the rate of change of the function with respect to its input variable. In this case, we are given the derivative of a function $f$, and we need to use this information to determine the critical points of $f$, the intervals on which $f$ is increasing or decreasing, and the points at which $f$ has a local maximum or minimum.

Q&A

Q: What are the critical points of $f$?

A: To find the critical points of $f$, we need to find the values of $x$ for which the derivative $f'(x)$ is equal to zero or undefined. In this case, the derivative is a polynomial function, so it is defined for all real numbers. Therefore, we need to find the values of $x$ for which $f'(x) = 0$.

Q: How do I find the critical points of a function?

A: To find the critical points of a function, you need to set the derivative equal to zero and solve for $x$. You can use the quadratic formula to solve the resulting equation.

Q: What if the derivative is not a polynomial function?

A: If the derivative is not a polynomial function, you may need to use other methods to find the critical points. For example, if the derivative is a rational function, you can use the method of partial fractions to find the critical points.

Q: How do I determine the intervals on which a function is increasing or decreasing?

A: To determine the intervals on which a function is increasing or decreasing, you need to examine the sign of the derivative. If the derivative is positive, the function is increasing. If the derivative is negative, the function is decreasing.

Q: What if the derivative is equal to zero?

A: If the derivative is equal to zero, the function may be increasing or decreasing on that interval. You need to examine the sign of the derivative on either side of the critical point to determine which of these is the case.

Q: How do I find the local maxima and minima of a function?

A: To find the local maxima and minima of a function, you need to examine the sign of the derivative on either side of the critical point. If the derivative changes from positive to negative, the function has a local maximum. If the derivative changes from negative to positive, the function has a local minimum.

Q: What if the function has no critical points?

A: If the function has no critical points, it may be increasing or decreasing on all intervals. You need to examine the sign of the derivative to determine which of these is the case.

Conclusion

In conclusion, the critical points of a function are the values of the input variable at which the derivative is equal to zero or undefined. The intervals on which a function is increasing or decreasing can be determined by examining the sign of the derivative. A function has a local maximum or minimum at a point where the function changes from increasing to decreasing or from decreasing to increasing, respectively.

Key Takeaways

• The critical points of a function are the values of the input variable at which the derivative is equal to zero or undefined.
• The intervals on which a function is increasing or decreasing can be determined by examining the sign of the derivative.
• A function has a local maximum or minimum at a point where the function changes from increasing to decreasing or from decreasing to increasing, respectively.

Further Reading

For further reading on this topic, we recommend the following resources:

• Calculus: Early Transcendentals
• Calculus: Single Variable
• Mathematics for Calculus

Practice Problems

To practice solving problems like this, we recommend the following exercises:

• Find the critical points of the function $f(x) = x^2 - 4x + 3$.
• Determine the intervals on which the function $f(x) = x^2 - 4x + 3$ is increasing or decreasing.
• Find the local maxima and minima of the function $f(x) = x^2 - 4x + 3$.