Critical Points and Relative Extrema

Oct 13, 2025 by miodeya.com 91 views

Understanding the Problem

Given the derivative of a function ${$ f^{\prime}(x) = \frac{x^2 - 8x}{(x-4)^2} $}$, we are tasked with finding the ${$ x $}$-value where a relative maximum exists for the original function ${$ f(x) $}$. This problem requires us to apply the concept of derivatives and critical points to determine the location of a relative maximum.

Critical Points and Relative Extrema

A critical point is a point on the function where the derivative is equal to zero or undefined. In this case, we are given the derivative ${$ f^{\prime}(x) = \frac{x^2 - 8x}{(x-4)^2} $}$. To find the critical points, we need to set the derivative equal to zero and solve for ${$ x $}$.

Finding Critical Points

To find the critical points, we set the derivative equal to zero:

${$ \frac{x^2 - 8x}{(x-4)^2} = 0 $}$

Since the denominator ${$ (x-4)^2 $}$ cannot be equal to zero, we can divide both sides of the equation by ${$ (x-4)^2 $}$:

${$ x^2 - 8x = 0 $}$

Now, we can factor out an ${$ x $}$ from the left-hand side of the equation:

${$ x(x - 8) = 0 $}$

This gives us two possible solutions for ${$ x $}$:

${$ x = 0 $}$ or ${$ x = 8 $}$

However, we need to check if these solutions are valid by plugging them back into the original derivative. If the derivative is undefined at either of these points, then they are not valid critical points.

Checking Validity of Critical Points

Let's check if the derivative is undefined at ${$ x = 0 $}$:

${$ f^{\prime}(0) = \frac{0^2 - 8(0)}{(0-4)^2} = \frac{0}{16} = 0 $}$

Since the derivative is defined at ${$ x = 0 $}$, this point is a valid critical point.

Now, let's check if the derivative is undefined at ${$ x = 8 $}$:

${$ f^{\prime}(8) = \frac{8^2 - 8(8)}{(8-4)^2} = \frac{64 - 64}{16} = \frac{0}{16} = 0 $}$

Since the derivative is defined at ${$ x = 8 $}$, this point is also a valid critical point.

Determining Relative Maximum

Now that we have found the critical points, we need to determine which one corresponds to a relative maximum. To do this, we can use the second derivative test.

Second Derivative Test

The second derivative of a function is the derivative of the first derivative. In this case, we need to find the second derivative of ${$ f^{\prime}(x) = \frac{x^2 - 8x}{(x-4)^2} $}$.

To find the second derivative, we can use the quotient rule:

${$ f^{\prime\prime}(x) = \frac{(x-4)^2(2x - 8) - (x^2 - 8x)(2(x-4))}{(x-4)^4} $}$

Simplifying the expression, we get:

${$ f^{\prime\prime}(x) = \frac{(x-4)(2x - 8) - 2(x^2 - 8x)}{(x-4)^3} $}$

Now, we can evaluate the second derivative at the critical points:

${$ f^{\prime\prime}(0) = \frac{(0-4)(2(0) - 8) - 2(0^2 - 8(0))}{(0-4)^3} = \frac{32}{64} = \frac{1}{2} $}$

Since the second derivative is positive at ${$ x = 0 $}$, this point corresponds to a relative minimum.

${$ f^{\prime\prime}(8) = \frac{(8-4)(2(8) - 8) - 2(8^2 - 8(8))}{(8-4)^3} = \frac{48}{192} = \frac{1}{4} $}$

Since the second derivative is positive at ${$ x = 8 $}$, this point also corresponds to a relative minimum.

Conclusion

In conclusion, we have found that the critical points ${$ x = 0 $}$ and ${$ x = 8 $}$ correspond to relative minima, not relative maxima. Therefore, the correct answer is not among the options provided.

However, if we re-examine the original derivative, we notice that the denominator ${$ (x-4)^2 $}$ is equal to zero when ${$ x = 4 $}$. This means that the derivative is undefined at ${$ x = 4 $}$, making it a critical point.

Re-evaluating the Second Derivative Test

Since the second derivative test is inconclusive at ${$ x = 4 $}$, we need to use other methods to determine the nature of the critical point.

One way to do this is to use the first derivative test. We can evaluate the first derivative at points near ${$ x = 4 $}$ to determine the sign of the derivative.

First Derivative Test

Let's evaluate the first derivative at ${$ x = 3 $}$ and ${$ x = 5 $}$:

${$ f^{\prime}(3) = \frac{3^2 - 8(3)}{(3-4)^2} = \frac{9 - 24}{1} = -15 $}$

Since the derivative is negative at ${$ x = 3 $}$, we know that the function is decreasing at this point.

${$ f^{\prime}(5) = \frac{5^2 - 8(5)}{(5-4)^2} = \frac{25 - 40}{1} = -15 $}$

Since the derivative is negative at ${$ x = 5 $}$, we know that the function is decreasing at this point.

Conclusion

Since the function is decreasing on both sides of ${$ x = 4 $}$, we can conclude that the function has a relative maximum at ${$ x = 4 $}$.

Therefore, the correct answer is:

$**A. ${$ x = 4 \$}$**<br/> **Q&A: Determining Relative Maximum Using Derivatives**$

Q: What is a relative maximum?

A: A relative maximum is a point on a function where the function changes from increasing to decreasing. It is a local maximum, meaning that it is the highest point in a small neighborhood of the point.

Q: How do you find the relative maximum of a function?

A: To find the relative maximum of a function, you need to find the critical points of the function. Critical points are points where the derivative of the function is equal to zero or undefined. You can then use the second derivative test to determine which critical points correspond to relative maxima.

Q: What is the second derivative test?

A: The second derivative test is a method used to determine the nature of a critical point. It involves evaluating the second derivative of the function at the critical point. If the second derivative is positive, the critical point corresponds to a relative minimum. If the second derivative is negative, the critical point corresponds to a relative maximum.

Q: What is the first derivative test?

A: The first derivative test is a method used to determine the nature of a critical point. It involves evaluating the first derivative of the function at points near the critical point. If the first derivative is negative on both sides of the critical point, the critical point corresponds to a relative maximum.

Q: How do you determine the nature of a critical point when the second derivative test is inconclusive?

A: When the second derivative test is inconclusive, you can use the first derivative test to determine the nature of the critical point. You can also use other methods, such as graphing the function or using numerical methods.

Q: What is the significance of the derivative being undefined at a critical point?

A: The derivative being undefined at a critical point means that the function has a vertical tangent line at that point. This can indicate that the function has a relative maximum or minimum at that point.

Q: Can a function have multiple relative maxima?

A: Yes, a function can have multiple relative maxima. This occurs when the function has multiple critical points that correspond to relative maxima.

Q: How do you determine the relative maximum of a function with multiple critical points?

A: To determine the relative maximum of a function with multiple critical points, you need to evaluate the second derivative at each critical point. If the second derivative is negative at a critical point, that point corresponds to a relative maximum.

Q: Can a function have a relative maximum at a point where the derivative is undefined?

A: Yes, a function can have a relative maximum at a point where the derivative is undefined. This occurs when the function has a vertical tangent line at that point.

Q: How do you determine the relative maximum of a function with a vertical tangent line?

A: To determine the relative maximum of a function with a vertical tangent line, you need to use the first derivative test. You can evaluate the first derivative at points near the point where the derivative is undefined. If the first derivative is negative on both sides of that point, the point corresponds to a relative maximum.

Q: Can a function have a relative maximum at a point where the second derivative is zero?

A: Yes, a function can have a relative maximum at a point where the second derivative is zero. This occurs when the second derivative test is inconclusive.

Q: How do you determine the relative maximum of a function with a second derivative of zero?

A: To determine the relative maximum of a function with a second derivative of zero, you need to use the first derivative test. You can evaluate the first derivative at points near the point where the second derivative is zero. If the first derivative is negative on both sides of that point, the point corresponds to a relative maximum.