How do you find the critical points for #f(x) = (x^2-10x)^4# and the local max and min?

Question

Aurora Alvarado · Accepted Answer

three critical points: #x=0,5,10#
Relative minimum at #x=0, 10#
Relative maximum at #x=5# First, we need to find #f'(x)#. Using rules for taking derivative, we get following: #f'(x) = 4*(x^2-10x)^3*(2x-10)# Critical points occur when the slope is zero meaning #f'(x)=0#. #f'(x) = 4*(x^2-10x)^3*(2x-10)=0# #(2x-10)=0# => #x=5# #(x^2-10x)=0# => #x=0, 10# So, three critical points would be #x=0,5,10#. To find relative maximum and minimum, we can do a sign test. For #x in#(-∞, 0) => #f'(x)# is negative. For #x in#(0, 5) => #f'(x)# is positive. For #x in#(5, 10) => #f'(x)# is negative. For #x in#(10, ∞) => #f'(x)# is positive. At #x=0#, function has relative minimum because #f'(x)# changes signs from negative to positive. At #x=5#, function has relative maximum because #f'(x)# changes signs from positive to negative. At #x=10#, function has relative minimum because #f'(x)# changes signs from negative to positive.

Paisley Johnson · Answer

undefined To find the critical points for $ f(x) = (x^2 - 10x)^4 $ and the local maximum and minimum, follow these steps:

1. Compute the first derivative of the function $ f(x) $.
2. Find the values of $ x $ where the derivative equals zero or is undefined.
3. Test the critical points by evaluating the second derivative at each point.
4. Determine the nature of each critical point:

a. If the second derivative is positive, it's a local minimum.
   b. If the second derivative is negative, it's a local maximum.
   c. If the second derivative is zero or undefined, the test is inconclusive.

5. Compute $ f(x) $ at each critical point to find the corresponding local maximum or minimum values.

Let's proceed with the steps:

1. Compute the first derivative of $ f(x) $:
$$ f'(x) = 4(x^2 - 10x)^3 \cdot (2x - 10) $$

2. Find the critical points by setting the first derivative equal to zero:
$$ 4(x^2 - 10x)^3 \cdot (2x - 10) = 0 $$

This gives us critical points at $ x = 0 $ and $ x = 5 $.

3. Test the critical points by evaluating the second derivative:
$$ f''(x) = 4 \cdot 3(x^2 - 10x)^2 \cdot (2x - 10) + 4(x^2 - 10x)^3 \cdot 2 $$

4. Evaluate $ f''(x) $ at $ x = 0 $ and $ x = 5 $:
   - At $ x = 0 $:
     - $ f''(0) = 4 \cdot 3(0 - 0)^2 \cdot (2 \cdot 0 - 10) + 4(0 - 0)^3 \cdot 2 = 0 $ (inconclusive)
   - At $ x = 5 $:
     - $ f''(5) = 4 \cdot 3(5^2 - 10 \cdot 5)^2 \cdot (2 \cdot 5 - 10) + 4(5^2 - 10 \cdot 5)^3 \cdot 2 > 0 $ (positive)

5. Determine the nature of each critical point:
   - At $ x = 0 $, the nature is inconclusive.
   - At $ x = 5 $, it's a local minimum.

6. Compute $ f(x) $ at $ x = 0 $ and $ x = 5 $:
   - At $ x = 0 $: $ f(0) = (0^2 - 10 \cdot 0)^4 = 0 $
   - At $ x = 5 $: $ f(5) = (5^2 - 10 \cdot 5)^4 = 25^4 $

Therefore, the critical points for $ f(x) $ are $ x = 0 $ and $ x = 5 $, and $ f(x) $ has a local minimum at $ x = 5 $.