Let $f(x) = -4x^3 + 6x^2 - 5$.The Absolute Minimum Value Of $f$ Over The Closed Interval $-2 \leq X \leq 3$ Occurs At What $ X X X [/tex]-value?Choose One Answer:A. 3 B. -2 C. 1 D. 0

Introduction

In calculus, finding the absolute minimum value of a function over a closed interval is a crucial problem. This involves identifying the critical points of the function, which are the points where the derivative of the function is zero or undefined. In this article, we will explore how to find the absolute minimum value of a cubic function over a closed interval.

The Function

The given function is $f(x) = -4x^3 + 6x^2 - 5$. This is a cubic function, which means it has a degree of 3. The function has a negative leading coefficient, indicating that the function is decreasing as $x$ increases.

The Closed Interval

The closed interval over which we want to find the absolute minimum value of the function is $-2 \leq x \leq 3$. This means that we need to consider the values of the function at the endpoints of the interval, as well as any critical points that may occur within the interval.

Finding Critical Points

To find the critical points of the function, we need to take the derivative of the function and set it equal to zero. The derivative of the function is given by:

$$f'(x) = -12x^2 + 12x$$

Setting the derivative equal to zero, we get:

$$-12x^2 + 12x = 0$$

Simplifying the equation, we get:

$$x^2 - x = 0$$

Factoring the equation, we get:

$$x(x - 1) = 0$$

This gives us two possible values for $x$: $x = 0$ and $x = 1$.

Evaluating the Function at Critical Points

Now that we have found the critical points, we need to evaluate the function at these points to determine which one corresponds to the absolute minimum value.

Evaluating the function at $x = 0$, we get:

$$f(0) = -4(0)^3 + 6(0)^2 - 5 = -5$$

Evaluating the function at $x = 1$, we get:

$$f(1) = -4(1)^3 + 6(1)^2 - 5 = -3$$

Evaluating the Function at Endpoints

We also need to evaluate the function at the endpoints of the interval, which are $x = -2$ and $x = 3$.

Evaluating the function at $x = -2$, we get:

$$f(-2) = -4(-2)^3 + 6(-2)^2 - 5 = -3$$

Evaluating the function at $x = 3$, we get:

$$f(3) = -4(3)^3 + 6(3)^2 - 5 = -91$$

Comparing Values

Now that we have evaluated the function at all the critical points and endpoints, we can compare the values to determine which one corresponds to the absolute minimum value.

The values we obtained are:

• $f(0) = -5$
• $f(1) = -3$
• $f(-2) = -3$
• $f(3) = -91$

Comparing these values, we see that the smallest value is $-91$, which occurs at $x = 3$.

Conclusion

In conclusion, the absolute minimum value of the function $f(x) = -4x^3 + 6x^2 - 5$ over the closed interval $-2 \leq x \leq 3$ occurs at $x = 3$.

Answer

The correct answer is:

A. 3

Discussion

This problem requires the application of calculus concepts, such as finding critical points and evaluating functions at these points. It also requires an understanding of the behavior of cubic functions and how to identify the absolute minimum value over a closed interval.

Additional Resources

For more information on calculus and finding absolute minimum values, see the following resources:

• Calculus Tutorial
• Absolute Minimum Value

References

• Calculus by Michael Spivak
• Calculus by James Stewart
  Q&A: Finding the Absolute Minimum Value of a Cubic Function ===========================================================

Q: What is the absolute minimum value of a function?

A: The absolute minimum value of a function is the smallest value that the function can take over a given interval. It is the minimum value that the function can achieve, and it occurs at a specific point or points within the interval.

Q: How do I find the absolute minimum value of a function?

A: To find the absolute minimum value of a function, you need to follow these steps:

1. Find the critical points of the function by taking the derivative and setting it equal to zero.
2. Evaluate the function at the critical points.
3. Evaluate the function at the endpoints of the interval.
4. Compare the values obtained in steps 2 and 3 to determine which one is the smallest.

Q: What is a critical point?

A: A critical point is a point where the derivative of the function is zero or undefined. It is a point where the function may have a maximum or minimum value.

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

A: To find the critical points of a function, you need to take the derivative of the function and set it equal to zero. You can then solve for the values of x that make the derivative equal to zero.

Q: What is the difference between a maximum and a minimum?

A: A maximum is the largest value that a function can take over a given interval. A minimum is the smallest value that a function can take over a given interval.

Q: Can a function have both a maximum and a minimum?

A: Yes, a function can have both a maximum and a minimum. This occurs when the function has multiple critical points, and the function is concave up or concave down at each critical point.

Q: How do I know if a function is concave up or concave down?

A: To determine if a function is concave up or concave down, you need to look at the second derivative of the function. If the second derivative is positive, the function is concave up. If the second derivative is negative, the function is concave down.

Q: What is the second derivative?

A: The second derivative is the derivative of the first derivative. It is used to determine the concavity of a function.

Q: Can I use a graphing calculator to find the absolute minimum value of a function?

A: Yes, you can use a graphing calculator to find the absolute minimum value of a function. You can graph the function and use the calculator to find the minimum value.

Q: What are some common mistakes to avoid when finding the absolute minimum value of a function?

A: Some common mistakes to avoid when finding the absolute minimum value of a function include:

• Not evaluating the function at the endpoints of the interval.
• Not evaluating the function at the critical points.
• Not comparing the values obtained in steps 2 and 3 to determine which one is the smallest.
• Not using the second derivative to determine the concavity of the function.

Q: How do I know if I have found the absolute minimum value of a function?

A: You know you have found the absolute minimum value of a function when you have evaluated the function at all the critical points and endpoints, and you have compared the values to determine which one is the smallest.

Conclusion

Finding the absolute minimum value of a function is an important concept in calculus. By following the steps outlined in this article, you can find the absolute minimum value of a function and understand the behavior of the function over a given interval.