Question

Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible ) whether they correspond to local maxima or local minima. $$f(x)=6 x^{4} \ln x^{2}-7 x^{4}$$

Short answer

Question: Determine the critical points of the function $$f(x) = 6x^{4}\ln{x^2} - 7x^{4}$$ and use the Second Derivative Test to classify them as local maxima, local minima, or neither. Answer: The only critical point is x = 0, but its nature cannot be determined using the Second Derivative Test.

Step-by-step solution

Calculate the first derivative of f(x)

Find the first derivative of the given function, $$f(x) = 6x^{4}\ln{x^2} - 7x^{4}$$ Using the product rule, we get $$f'(x) = (6x^4)' (\ln(x^2))+ (6x^4)(\ln(x^2))' - (7x^4)'$$ Now, compute the derivatives of each term separately: $$ (6x^4)' = 24x^3 $$ $$ (\ln(x^2))' = \frac{1}{x^2} \cdot 2x $$ $$ (7x^4)' = 28x^3 $$ Substitute these back into the expression for $$f'(x)$$: $$f'(x) = 24x^3(\ln(x^2)) + 6x^4\left(\frac{2x}{x^2}\right) - 28x^3$$ Simplify and rewrite $$f'(x)$$: $$f'(x) = 24x^3\ln(x^2) + 12x^2 - 28x^3$$

Find the critical points

To find the critical points, set the first derivative equal to zero: $$ 24x^3\ln(x^2) + 12x^2 - 28x^3 = 0 $$ First, take out the factor of x^2: $$ x^2(24x\ln(x^2) + 12 - 28x) = 0 $$ Therefore, x = 0 is one critical point. Now, to find the other critical points, set the expression in the brackets equal to zero: $$ 24x\ln(x^2) + 12 - 28x = 0 $$ We cannot find any way to simplify this equation further, so we will find the second derivative first.

Calculate the second derivative of f(x)

Find the second derivative by differentiating the first derivative with respect to x: $$f''(x) = \frac{d}{dx}(24x^3\ln(x^2) + 12x^2 - 28x^3)$$ Now, compute the derivatives of each term separately using the sum and product rules: $$ (24x^3\ln(x^2))''' = 24(3x^2\ln(x^2) + x^3\cdot\frac{4x}{x^2}) = 72x^2\ln(x^2) + 96x^2 $$ $$ (12x^2)''' = 24x $$ $$ (28x^3)''' = 84x^2 $$ Substitute these back into the expression for $$f''(x)$$: $$f''(x) = 72x^2\ln(x^2) + 96x^2 + 24x - 84x^2$$ Simplify and rewrite $$f''(x)$$: $$f''(x) = 72x^2\ln(x^2) + 12x^2 + 24x$$

Apply the Second Derivative Test

Use the second derivative test to classify the critical points: For x = 0: $$f''(0) = 72(0)^2\ln(0^2) + 12(0)^2 + 24(0)$$ The $$\ln(0^2)$$ term is undefined, so we cannot conclude anything about x=0 using the Second Derivative Test. Now, let's consider the expression, $$ 24x\ln(x^2) + 12 - 28x = 0 $$ Using numerical methods or graphing, we would find that this equation does not have real roots. So, x = 0 is the only critical point, but its nature cannot be determined using the Second Derivative Test.

Key concepts

First Derivative

The first derivative of a function is crucial because it tells us how the function behaves at different points. By finding where the first derivative equals zero or does not exist, we identify critical points. These are the points where the function could potentially have local maxima or minima.

To calculate the first derivative of a complex function like \( f(x) = 6x^{4} \ln{x^2} - 7x^{4} \), we often use the product rule, especially when the function is a product of two parts. In this case:

• The term \( (6x^4)' \) differentiates to \( 24x^3 \),
• \( (\ln(x^2))' \) becomes \( \frac{1}{x^2} \cdot 2x \),
• and finally, \( (7x^4)' \) is simplified to \( 28x^3 \).

These derivatives help simplify the original expression to find potential zeros of the derivative, which indicate critical points.

The critical point in this example, found by setting \( f'(x) = 0 \), happens to be at \( x = 0 \), telling us that the function could change its increasing or decreasing nature here.

Second Derivative

After using the first derivative to locate critical points, the second derivative of a function aids in understanding the concavity of the function. This tells us if the function is curving upwards or downwards at those points.

In this exercise, the second derivative is derived from simplifying the expression created by further differentiating \( f'(x) \). This involves breaking down each part once again:

• \( (24x^3 \ln(x^2))'' \) becomes \( 72x^2 \ln(x^2) + 96x^2 \)
• \( (12x^2)'' \) results in \( 24x \)
• \( (28x^3)'' \) equals \( 84x^2 \)

The complete second derivative \( f''(x) \) is then simplified to \( 72x^2 \ln(x^2) + 12x^2 + 24x \). This formula helps indicate how the graph behaves at and around the critical points, although special cases like undefined logarithms can occur.

Second Derivative Test

The Second Derivative Test is a method for classifying critical points as local minima or maxima. It is practical when the second derivative is easy to calculate and evaluate.

For a critical point \( x_0 \), if:

• \( f''(x_0) > 0 \), the function has a local minimum at \( x_0 \).
• \( f''(x_0) < 0 \), the function has a local maximum at \( x_0 \).
• \( f''(x_0) = 0 \), the test is inconclusive.

In the exercise, we attempted to apply this test to the critical point \( x = 0 \) using the second derivative \( f''(x) \). However, evaluating \( f''(0) \) leads to undefined terms because of \( \ln(0^2) \), making the test inconclusive. Therefore, sometimes other methods must be used to fully understand the nature of critical points.