Question

let f be a function with a second derivative given by $f(x) = x^2(x - 3)(x - 6)$. what are the x-coordinates of the points of inflection of the graph of f?

a 0 only

b 3 only

c 0 and 6 only

d 3 and 6 only

e 0, 3, and 6

Explanation:

Step1: Recall Inflection Point Rule

A point of inflection occurs where the second derivative \( f''(x) \) changes sign (i.e., the concavity changes), and \( f''(x) = 0 \) or is undefined (here \( f''(x) \) is a polynomial, so defined everywhere). First, find critical points by solving \( f''(x)=0 \).

Given \( f''(x)=x^2(x - 3)(x - 6) \), set to zero:
\( x^2(x - 3)(x - 6)=0 \)
Solutions: \( x = 0 \), \( x = 3 \), \( x = 6 \) (since \( x^2 = 0 \Rightarrow x = 0 \), \( x - 3 = 0 \Rightarrow x = 3 \), \( x - 6 = 0 \Rightarrow x = 6 \)).

Step2: Test Sign Changes of \( f''(x) \)

• For \( x = 0 \):

Analyze intervals around \( x = 0 \), e.g., \( x=-1 \) (left of 0): \( (-1)^2(-1 - 3)(-1 - 6)=1(-4)(-7)=28>0 \).
\( x = 0.5 \) (right of 0, left of 3): \( (0.5)^2(0.5 - 3)(0.5 - 6)=0.25(-2.5)(-5.5)=3.4375>0 \).
Sign of \( f''(x) \) does not change at \( x = 0 \) (both sides positive). So \( x = 0 \) is not an inflection point (since concavity doesn’t change).

• For \( x = 3 \):

Intervals: \( x = 2 \) (left of 3): \( 2^2(2 - 3)(2 - 6)=4(-1)(-4)=16>0 \).
\( x = 4 \) (right of 3, left of 6): \( 4^2(4 - 3)(4 - 6)=16(1)(-2)=-32<0 \).
Sign changes from positive to negative: concavity changes (inflection point at \( x = 3 \)).

• For \( x = 6 \):

Intervals: \( x = 5 \) (left of 6): \( 5^2(5 - 3)(5 - 6)=25(2)(-1)=-50<0 \).
\( x = 7 \) (right of 6): \( 7^2(7 - 3)(7 - 6)=49(4)(1)=196>0 \).
Sign changes from negative to positive: concavity changes (inflection point at \( x = 6 \)).

Answer:

D. 3 and 6 only