Question

the second derivative of a function $f$ is given by $f(x) = x(x - 3)^5(x - 10)^2$. at which of the following values of $x$ does the graph of $f$ have a point of inflection?
a 3 only
b 0 and 3 only
c 3 and 10 only
d 0, 3, and 10

Explanation:

Step1: Recall Inflection Point Rule

A point of inflection occurs where \( f''(x) = 0 \) and \( f''(x) \) changes sign (concavity changes) around that point.

Step2: Analyze \( f''(x) = x(x - 3)^5(x - 10)^2 \)

- **For \( x = 0 \):**
The factor \( x \) is linear. For \( x < 0 \), \( x < 0 \); for \( x > 0 \), \( x > 0 \). But check the other factors: \( (x - 3)^5 \) (for \( x \) near 0, \( x - 3 < 0 \), so \( (x - 3)^5 < 0 \)) and \( (x - 10)^2 > 0 \) (square is non - negative). So \( f''(x) \) near \( x = 0 \): left of 0: \( (-)(-)(+) = + \); right of 0: \( (+)(-)(+) = - \). Wait, no—wait, actually, when \( x \) is just less than 0, \( x=- \epsilon \) (\( \epsilon>0 \)), \( (x - 3)=-3 - \epsilon\), so \( (x - 3)^5=(-)^5=- \), \( (x - 10)^2=( - 10-\epsilon)^2 = + \). So \( f''(-\epsilon)=(-\epsilon)(-)(+)=+\epsilon>0 \). When \( x \) is just greater than 0, \( x = \epsilon \), \( (x - 3)=-3+\epsilon\approx - 3\), \( (x - 3)^5=- \), \( (x - 10)^2=( - 10+\epsilon)^2 = + \). So \( f''(\epsilon)=(\epsilon)(-)(+)=-\epsilon<0 \). Wait, but wait—actually, we need to check the multiplicity of the roots. Wait, no, the key is: a root of \( f''(x) \) with odd multiplicity will cause a sign change, even multiplicity will not.
- \( x = 0 \): multiplicity 1 (odd)
- \( x = 3 \): multiplicity 5 (odd)
- \( x = 10 \): multiplicity 2 (even)

Step3: Check Sign Change at Each Root

- **\( x = 0 \):**
As above, \( f''(x) \) changes sign (from + to -) around \( x = 0 \)? Wait, no—wait, when \( x \) is near 0, left: \( x<0 \), \( (x - 3)<0 \), \( (x - 10)^2>0 \), so \( f''(x)=x(x - 3)^5(x - 10)^2 \): \( (-)(-)(+)=+ \). Right: \( x>0 \), \( (x - 3)<0 \), \( (x - 10)^2>0 \), so \( f''(x)=(+)(-)(+)=- \). So sign changes at \( x = 0 \)? Wait, but wait, let's check \( x = 3 \):
For \( x \) near 3, left of 3: \( x = 3-\epsilon \), \( x>0 \) (if \( \epsilon < 3 \)), \( (x - 3)=-\epsilon \), \( (x - 10)^2=( - 7-\epsilon)^2>0 \). So \( f''(3 - \epsilon)=(3 - \epsilon)(-\epsilon)^5(positive) \). Since \( (-\epsilon)^5=- \epsilon^5 \), so \( f''(3 - \epsilon)=(positive)(-)(positive)=- \). Right of 3: \( x = 3+\epsilon \), \( (x - 3)=\epsilon \), \( (x - 3)^5=\epsilon^5 \), so \( f''(3+\epsilon)=(3+\epsilon)(\epsilon^5)(positive)=+ \). So sign changes at \( x = 3 \).

For \( x = 10 \): \( (x - 10)^2 \) is a square, so regardless of \( x \) slightly less than 10 or slightly more than 10, \( (x - 10)^2 \) is positive. The other factors: \( x \) (for \( x \) near 10, \( x>0 \)) and \( (x - 3)^5 \) (for \( x \) near 10, \( x - 3>0 \), so \( (x - 3)^5>0 \)). So \( f''(x) \) near 10: left of 10 (\( x = 10-\epsilon \)): \( (10 - \epsilon)(7 - \epsilon)^5(\epsilon)^2 \), all positive (since \( 10-\epsilon>0 \), \( 7 - \epsilon>0 \), \( \epsilon^2>0 \)). Right of 10 (\( x = 10+\epsilon \)): \( (10+\epsilon)(7+\epsilon)^5(\epsilon)^2 \), all positive. So \( f''(x) \) does not change sign at \( x = 10 \) (multiplicity 2, even).

Wait, but earlier analysis for \( x = 0 \): let's re - check. Wait, when \( x = 0 \), the factor \( x \) is linear. But let's take \( x=-1 \) (left of 0): \( f''(-1)=(-1)(-4)^5(-11)^2=(-1)(-1024)(121)=123704>0 \). \( x = 1 \) (right of 0): \( f''(1)=(1)(-2)^5(-9)^2=(1)(-32)(81)=-2592<0 \). So sign changes at \( x = 0 \). Wait, but the answer options: option B is 0 and 3 only, option A is 3 only. Wait, what's wrong here? Wait, no—wait, the definition of inflection point: a point where the concavity changes, i.e., \( f''(x) \) changes sign. But let's check the multiplicity again. Wait, the formula for \( f''(x)=x(x - 3)^5(x - 10)^2 \). The roots are at \( x = 0 \) (multiplicity 1), \( x = 3 \) (multiplicity 5), \( x = 10 \) (multiplicity 2). For a root of \( f''(x) \), if the multiplicity is odd, \( f''(x) \) changes sign; if even, it doesn't. So \( x = 0 \) (multiplicity 1, odd) – sign changes; \( x = 3 \) (multiplicity 5, odd) – sign changes; \( x = 10 \) (multiplicity 2, even) – no sign change. Wait, but when we calculated \( f''(-1) \) and \( f''(1) \), we saw sign change at \( x = 0 \). But the answer options: option B is 0 and 3 only, but let's check with the standard method. Wait, maybe I made a mistake. Wait, let's take \( x = 0 \): the function \( f''(x) \) at \( x = 0 \) is 0. Now, check concavity: for \( x < 0 \), say \( x=-1 \), \( f''(-1)=(-1)(-4)^5(-11)^2=(-1)(-1024)(121)=123704>0 \) (concave up). For \( x > 0 \), say \( x = 1 \), \( f''(1)=(1)(-2)^5(-9)^2=(1)(-32)(81)=-2592<0 \) (concave down). So concavity changes at \( x = 0 \), so inflection point at \( x = 0 \). At \( x = 3 \): \( x = 2 \) (left of 3): \( f''(2)=(2)(-1)^5(-8)^2=(2)(-1)(64)=-128<0 \) (concave down). \( x = 4 \) (right of 3): \( f''(4)=(4)(1)^5(-6)^2=(4)(1)(36)=144>0 \) (concave up). So concavity changes at \( x = 3 \). At \( x = 10 \): \( x = 9 \) (left of 10): \( f''(9)=(9)(6)^5(-1)^2=(9)(7776)(1)=69984>0 \) (concave up). \( x = 11 \) (right of 10): \( f''(11)=(11)(8)^5(1)^2=(11)(32768)(1)=360448>0 \) (concave up). So no sign change at \( x = 10 \), so no inflection point. Wait, but the answer options: option B is 0 and 3 only. But wait, in the initial analysis, I thought \( x = 0 \) has sign change, \( x = 3 \) has sign change, \( x = 10 \) does not. So the inflection points are at \( x = 0 \) and \( x = 3 \)? But wait, let's check the original problem again. Wait, maybe I made a mistake in the sign for \( x = 0 \). Wait, \( f''(x)=x(x - 3)^5(x - 10)^2 \). At \( x=-1 \): \( (-1)(-4)^5(-11)^2=(-1)(-1024)(121)=123704>0 \) (concave up). At \( x = 1 \): \( (1)(-2)^5(-9)^2=(1)(-32)(81)=-2592<0 \) (concave down). So concavity changes at \( x = 0 \). At \( x = 3 \): \( x = 2 \): \( (2)(-1)^5(-8)^2=(2)(-1)(64)=-128<0 \) (concave down). \( x = 4 \): \( (4)(1)^5(-6)^2=(4)(1)(36)=144>0 \) (concave up). So concavity changes at \( x = 3 \). At \( x = 10 \): \( x = 9 \): \( (9)(6)^5(1)^2=9*7776=69984>0 \) (concave up). \( x = 11 \): \( (11)(8)^5(1)^2=11*32768=360448>0 \) (concave up). So no change at \( x = 10 \). So the inflection points are at \( x = 0 \) and \( x = 3 \). But wait, the answer option B is "0 and 3 only". But wait, let's check the multiplicity again. The root at \( x = 0 \) has multiplicity 1 (odd), so sign changes. Root at \( x = 3 \) has multiplicity 5 (odd), sign changes. Root at \( x = 10 \) has multiplicity 2 (even), no sign change. So the points of inflection are at \( x = 0 \) and \( x = 3 \). Wait, but the first analysis I thought maybe \( x = 0 \) is included, but let's check the answer options. Option B is 0 and 3 only. But wait, maybe I made a mistake. Wait, let's check the definition again: a point of inflection is a point where the concavity changes, i.e., \( f''(x) \) changes sign. So at \( x = 0 \), \( f''(x) \) changes from positive to negative (concavity from up to down), so inflection point. At \( x = 3 \), \( f''(x) \) changes from negative to positive (concavity from down to up), so inflection point. At \( x = 10 \), \( f''(x) \) is positive on both sides, so no inflection point. So the correct answer is B: 0 and 3 only.

Answer:

B. 0 and 3 only