Math Problem Statement

Can I use second order derivative here

Solution

Yes, you can use the second-order derivative to determine whether the critical points are local maxima, minima, or points of inflection. This method is known as the Second Derivative Test.

Here's how it works:

Step 1: Take the first derivative (which we've already done)

We have the first derivative:

f(x)=6x230x+36f'(x) = 6x^2 - 30x + 36

Step 2: Take the second derivative

Now, take the derivative of f(x)f'(x) to get the second derivative:

f(x)=ddx(6x230x+36)=12x30f''(x) = \frac{d}{dx}(6x^2 - 30x + 36) = 12x - 30

Step 3: Evaluate the second derivative at the critical points

We previously found the critical points to be x=2x = 2 and x=3x = 3. Now, let's plug these values into the second derivative:

  1. For x=2x = 2:

f(2)=12(2)30=2430=6f''(2) = 12(2) - 30 = 24 - 30 = -6

Since f(2)<0f''(2) < 0, this indicates that f(x)f(x) has a local maximum at x=2x = 2.

  1. For x=3x = 3:

f(3)=12(3)30=3630=6f''(3) = 12(3) - 30 = 36 - 30 = 6

Since f(3)>0f''(3) > 0, this indicates that f(x)f(x) has a local minimum at x=3x = 3.

Step 4: Compare the critical points with the endpoints

To determine the absolute maxima and minima on the interval [1,5][1, 5], we still need to compare the function values at the critical points x=2x = 2, x=3x = 3, and the endpoints x=1x = 1 and x=5x = 5 (as we did before).

  • f(1)=24f(1) = 24
  • f(2)=29f(2) = 29 (local maximum)
  • f(3)=28f(3) = 28 (local minimum)
  • f(5)=56f(5) = 56

Conclusion

  • The absolute maximum on [1,5][1, 5] is at x=5x = 5 with f(5)=56f(5) = 56.
  • The absolute minimum on [1,5][1, 5] is at x=1x = 1 with f(1)=24f(1) = 24.

The second derivative test helped us confirm the nature of the critical points, but we still needed to evaluate the endpoints to find the absolute extrema.

Would you like to explore more details or have any further questions?

Here are 5 questions to consider:

  1. What does it mean if the second derivative at a critical point is zero?
  2. How would the process change if the interval was open instead of closed?
  3. Can the second derivative test always determine the nature of a critical point?
  4. What if the second derivative doesn't exist at a critical point?
  5. How does concavity relate to the second derivative and extrema?

Tip: The second derivative test is a quick way to assess the nature of critical points, but don't forget to consider endpoints for absolute extrema on closed intervals.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives

Formulas

Second Derivative Test

Theorems

Second Derivative Test

Suitable Grade Level

College Level