How do you find the inflection points of the graph of the function: #f(x) = (x+2) (x-4)^2#?

Question

Samantha Adkison · Accepted Answer

#x=2# Considering function, #f(x)=(x+2)(x-4)^2# #f(x)=x^3-6x^2+32# #f'(x)=3x^2-12x# #f''(x)=6x-12# For, inflection point #f''(x)=0# # 6x-12=0# #x=2#

Elijah Abram · Answer

undefined To find the inflection points of the function $ f(x) = (x+2)(x-4)^2 $, you need to find the points where the concavity changes. Inflection points occur where the second derivative changes sign or where the second derivative is equal to zero.

1. Find the first derivative of the function.
   $ f'(x) = (x-4)^2 + (x+2)(2)(x-4) $

2. Find the second derivative of the function.
   $ f''(x) = 2(x-4) + (x+2)(2) + 2(x+2) $

3. Set the second derivative equal to zero and solve for $ x $.
   $ 2(x-4) + 2(x+2) + 2(x+2) = 0 $

4. Solve for $ x $ to find the $ x $-coordinate of the inflection point(s).

5. Once you find the $ x $-coordinate(s) of the potential inflection point(s), test the concavity around those points by examining the signs of the second derivative.

6. Confirm the concavity changes at those points by checking if the sign of the second derivative changes.

7. The point(s) where the concavity changes are the inflection point(s) of the function.