Question

9 mark for review the second derivative of the function f is given by $f(x) = x(x - a)(x - b)^2$. the graph of $f$ is shown above. for what values of x does the graph of f have a point of inflection? a 0 and a only b 0 and m only c b and j only d 0, a, and b e b, j, and k

Explanation:

Step1: Recall Inflection Point Rule

A point of inflection for \( f(x) \) occurs where \( f''(x) \) changes sign (concavity changes) and \( f''(x) = 0 \) or is undefined (here \( f''(x) \) is a polynomial, so defined everywhere).

Step2: Analyze \( f''(x) = x(x - a)(x - b)^2 \)

• Roots at \( x = 0 \), \( x = a \), \( x = b \) (with multiplicity 2 at \( x = b \)).
• For \( x = b \): Since \( (x - b)^2 \) is a square, the sign of \( f''(x) \) near \( x = b \) (left and right) will not change (because squaring preserves sign). So \( f''(x) \) does not change sign at \( x = b \), no inflection here.
• For \( x = 0 \) and \( x = a \): Check sign changes. From the graph, \( f''(x) \) crosses the x - axis at \( x = a \) and \( x = 0 \) (indicating sign change), while at \( x = b \) it touches the axis (no sign change). Wait, but the graph labels: Wait, the graph of \( f''(x) \) has roots at \( x = a \), \( x = 0 \), \( x = b \). Wait, maybe mislabeling: Wait, the option has 0 and a? Wait no, wait the options: Option A is 0 and a, but let's re - evaluate. Wait, the graph: the left root is \( x = a \), then \( x = 0 \), then \( x = b \). Wait, maybe the variable names: Wait, the function is \( f''(x)=x(x - a)(x - b)^2 \). So when \( x < a \), let's pick a test point, say \( x=a - 1 \), \( f''(a - 1)=(a - 1)(- 1)(a - 1 - b)^2 \). Between \( a \) and \( 0 \), say \( x=\frac{a}{2} \), \( f''(\frac{a}{2})=\frac{a}{2}(\frac{a}{2}-a)(\frac{a}{2}-b)^2=\frac{a}{2}(-\frac{a}{2})(\frac{a}{2}-b)^2 \). Between \( 0 \) and \( b \), say \( x=\frac{b}{2} \), \( f''(\frac{b}{2})=\frac{b}{2}(\frac{b}{2}-a)(\frac{b}{2}-b)^2=\frac{b}{2}(\frac{b}{2}-a)(-\frac{b}{2})^2 \). After \( b \), \( x = b + 1 \), \( f''(b + 1)=(b + 1)(b + 1 - a)(1)^2 \). But from the graph, the key is sign change. At \( x = a \): \( f''(x) \) goes from negative to positive (or vice - versa) as we cross \( x = a \). At \( x = 0 \): \( f''(x) \) crosses the axis, so sign change. At \( x = b \): since it's a double root, the graph touches the x - axis, so no sign change (concavity doesn't change). Wait, but the options: Option A is 0 and a, but let's check the options again. Wait, maybe I made a mistake. Wait the options are:

A. 0 and a only

B. 0 and m only

C. b and j only

D. 0, a, and b

E. b, j, and k

Wait, maybe the graph's labels: the points on the x - axis are \( x = a \), \( x = 0 \), \( x = b \). But the option A is 0 and a. Wait, but let's think again: A point of inflection requires \( f''(x) \) to change sign. For \( f''(x)=x(x - a)(x - b)^2 \), the factor \( (x - b)^2 \) is non - negative for all x. So the sign of \( f''(x) \) is determined by \( x(x - a) \). The product \( x(x - a) \) changes sign at \( x = 0 \) and \( x = a \) (because when \( x < a \), if \( x < 0 \), \( x(x - a)>0 \) (negative times negative), when \( 0 < x < a \), \( x(x - a)<0 \) (positive times negative), when \( x > a \), \( x(x - a)>0 \) (positive times positive)). The factor \( (x - b)^2 \) doesn't change the sign (just makes it zero at \( x = b \) and non - negative elsewhere). So \( f''(x) \) changes sign at \( x = 0 \) and \( x = a \), so inflection points at \( x = 0 \) and \( x = a \).

Answer:

A. 0 and a only