How do you find the critical numbers for #f(x)= x^(-8) ln x# to determine the maximum and minimum?

Question

How do you find the critical numbers for #f(x)= x^(-8)  ln x# to determine the maximum and minimum?

Emilia Aguilar · Accepted Answer

The function has no critical numbers.

A critical number for #f# is a number #c# in he domain of #f# where #f'(c) = 0# or #f'(c)# does not exist. #f(x) = x^-8 = 1/x^8# has domain: all reals except #0#. #f'(x) = -8x^-7# is never #0# and fails to exist only at #0#, which is not in the domain of #f#. So, #f# has no critical numbers.

Aurora Alvarado · Answer

undefined To find the critical numbers for $ f(x) = x^{-8} \ln(x) $ to determine the maximum and minimum, follow these steps:

1. Find the first derivative of $ f(x) $.
2. Set the derivative equal to zero and solve for $ x $.
3. Check for critical points where the derivative is undefined.
4. Evaluate the second derivative to determine concavity.
5. Determine the nature of the critical points using the second derivative test.

Let's proceed with these steps:

1. Find the first derivative:
$$ f'(x) = \frac{d}{dx} \left( x^{-8} \ln(x) \right) $$

2. Set the derivative equal to zero and solve for $ x $:
$$ f'(x) = 0 $$

3. Check for critical points where the derivative is undefined. $ f'(x) $ is undefined when $ x = 0 $, but $ x = 0 $ is not in the domain of $ f(x) $, so it's not a critical point.

4. Evaluate the second derivative:
$$ f''(x) = \frac{d^2}{dx^2} \left( x^{-8} \ln(x) \right) $$

5. Determine the nature of critical points using the second derivative test.

If $ f''(x) > 0 $, the function is concave up at that point, indicating a local minimum. 
If $ f''(x) < 0 $, the function is concave down at that point, indicating a local maximum.

You can now proceed with finding the first and second derivatives and applying the second derivative test to determine whether the critical points are maximum, minimum, or neither.