Question

7 mark for review
refer to the graph and the information below.
the function f is defined on the closed interval 0,8. the graph of its derivative f’ is shown above.
how many points of inflection does the graph of f have?
a two
b three
c four
d five
e six

Explanation:

Step1: Recall Inflection Point Rule

A point of inflection on \( f(x) \) occurs where \( f''(x) \) changes sign (i.e., \( f'(x) \) changes from increasing to decreasing or vice - versa, so the graph of \( f'(x) \) has a local maximum or minimum).

Step2: Analyze \( y = f'(x) \) Graph

- Look for local maxima and minima of \( y = f'(x) \).
- A local maximum or minimum of \( f'(x) \) corresponds to a point where \( f''(x)=0 \) and \( f''(x) \) changes sign (since at local extrema of \( f'(x) \), the slope of \( f'(x) \) (which is \( f''(x) \)) changes from positive to negative or negative to positive).
- From the graph of \( y = f'(x) \):
- At \( x = 1 \): \( f'(x) \) has a local maximum. So \( f''(x) \) changes from positive to negative here, which is an inflection point for \( f(x) \).
- At \( x = 3 \): \( f'(x) \) has a local maximum. So \( f''(x) \) changes from positive to negative here, which is an inflection point for \( f(x) \).
- At \( x = 5 \): \( f'(x) \) has a local maximum. So \( f''(x) \) changes from positive to negative here, which is an inflection point for \( f(x) \). Wait, no, wait. Wait, actually, we need to look for where \( f'(x) \) changes from increasing to decreasing or decreasing to increasing. Wait, let's re - examine:
- The local maxima of \( f'(x) \) are at \( x = 1 \), \( x = 3 \), \( x = 5 \)? No, wait the graph: Let's see the critical points of \( f'(x) \) (where \( f''(x)=0 \)) which are the local maxima and minima of \( f'(x) \).
- The local maxima of \( f'(x) \) occur at \( x = 1 \), \( x = 3 \), \( x = 5 \)? No, wait the graph: At \( x = 1 \), it's a peak (local max), at \( x = 3 \), it's a peak (local max), at \( x = 5 \), it's a peak (local max)? No, wait no, the local minima of \( f'(x) \) are at \( x = 2 \), \( x = 4 \), \( x = 7 \). Wait, no, let's think again. A point of inflection for \( f(x) \) is where \( f''(x) \) changes sign, which is where \( f'(x) \) changes from increasing to decreasing or decreasing to increasing (i.e., where \( f'(x) \) has a local maximum or minimum).
- Looking at the graph of \( y = f'(x) \):
- At \( x = 2 \): \( f'(x) \) has a local minimum (since it was increasing before \( x = 2 \) and decreasing after? Wait no, before \( x = 2 \), \( f'(x) \) was increasing to \( x = 1 \), then decreasing to \( x = 2 \) (a local min at \( x = 2 \)), then increasing to \( x = 3 \) (local max at \( x = 3 \)), then decreasing to \( x = 4 \) (local min at \( x = 4 \)), then increasing to \( x = 5 \) (local max at \( x = 5 \)), then decreasing to \( x = 7 \) (local min at \( x = 7 \)).
- So the local maxima of \( f'(x) \) are at \( x = 1 \), \( x = 3 \), \( x = 5 \) and local minima at \( x = 2 \), \( x = 4 \), \( x = 7 \). But for \( f(x) \) to have an inflection point, \( f''(x) \) must change sign, which happens when \( f'(x) \) changes from increasing to decreasing (local max of \( f'(x) \)) or decreasing to increasing (local min of \( f'(x) \)).
- At \( x = 2 \): \( f'(x) \) changes from decreasing to increasing (local min), so \( f''(x) \) changes from negative to positive, so this is an inflection point.
- At \( x = 4 \): \( f'(x) \) changes from decreasing to increasing (local min), so \( f''(x) \) changes from negative to positive, so this is an inflection point.
- At \( x = 7 \): \( f'(x) \) changes from decreasing to increasing (local min), so \( f''(x) \) changes from negative to positive, so this is an inflection point. Wait, no, I'm getting confused. Wait the correct rule: A point \( c \) in the domain of \( f \) is an inflection point if \( f''(c) = 0 \) (or \( f''(c) \) does not exist) and \( f''(x) \) changes sign at \( x = c \). Since \( f' \) is differentiable (the graph of \( f' \) is smooth), \( f''(x) \) exists, so we look for where \( f'(x) \) has a local maximum or minimum (since at those points, \( f''(x)=0 \) and \( f''(x) \) changes sign).
- Let's list the local extrema of \( f'(x) \):
- Local maxima of \( f'(x) \): at \( x = 1 \), \( x = 3 \), \( x = 5 \) (where \( f'(x) \) stops increasing and starts decreasing, so \( f''(x) \) changes from positive to negative).
- Local minima of \( f'(x) \): at \( x = 2 \), \( x = 4 \), \( x = 7 \) (where \( f'(x) \) stops decreasing and starts increasing, so \( f''(x) \) changes from negative to positive).
- But wait, the question is about the number of inflection points of \( f(x) \) on \([0,8]\). Wait, but let's check the graph again. Wait the original graph: Let's count the number of times \( f'(x) \) changes from increasing to decreasing or decreasing to increasing.
- At \( x = 2 \): \( f'(x) \) changes from decreasing to increasing (local min), so \( f''(x) \) changes from negative to positive: inflection point.
- At \( x = 4 \): \( f'(x) \) changes from decreasing to increasing (local min), so \( f''(x) \) changes from negative to positive: inflection point.
- At \( x = 1 \): \( f'(x) \) changes from increasing to decreasing (local max), so \( f''(x) \) changes from positive to negative: inflection point.
- At \( x = 3 \): \( f'(x) \) changes from increasing to decreasing (local max), so \( f''(x) \) changes from positive to negative: inflection point.
- At \( x = 5 \): \( f'(x) \) changes from increasing to decreasing (local max), so \( f''(x) \) changes from positive to negative: inflection point.
- At \( x = 7 \): \( f'(x) \) changes from decreasing to increasing (local min), so \( f''(x) \) changes from negative to positive: inflection point. Wait, no, this can't be. Wait, the answer options are Two, Three, Four, Five, Six. Wait, maybe I made a mistake. Wait, the key is that a point of inflection for \( f(x) \) is where \( f''(x) \) changes sign, which is where \( f'(x) \) has a local maximum or minimum. But let's look at the graph of \( f'(x) \):
- The local maxima of \( f'(x) \) are at \( x = 1 \), \( x = 3 \), \( x = 5 \) (3 points) and local minima at \( x = 2 \), \( x = 4 \), \( x = 7 \) (3 points). But wait, the function \( f(x) \) is defined on \([0,8]\). Wait, but maybe some of these points are not in the interval or maybe my analysis is wrong. Wait, let's recall: The graph of \( f'(x) \) is given. Let's look at the critical points of \( f'(x) \) (where \( f''(x) = 0 \)):
- The local maxima of \( f'(x) \) are at \( x = 1 \), \( x = 3 \), \( x = 5 \) (3 points) and local minima at \( x = 2 \), \( x = 4 \), \( x = 7 \) (3 points). But a point of inflection for \( f(x) \) occurs when \( f''(x) \) changes sign, which is when \( f'(x) \) changes from increasing to decreasing or decreasing to increasing.
- Wait, let's look at the graph again. Let's see the slope of \( f'(x) \) (which is \( f''(x) \)):
- From \( x = 0 \) to \( x = 1 \): \( f'(x) \) is increasing, so \( f''(x)>0 \).
- From \( x = 1 \) to \( x = 2 \): \( f'(x) \) is decreasing, so \( f''(x)<0 \). So at \( x = 1 \), \( f''(x) \) changes from positive to negative: inflection point.
- From \( x = 2 \) to \( x = 3 \): \( f'(x) \) is increasing, so \( f''(x)>0 \). So at \( x = 2 \), \( f''(x) \) changes from negative to positive: inflection point.
- From \( x = 3 \) to \( x = 4 \): \( f'(x) \) is decreasing, so \( f''(x)<0 \). So at \( x = 3 \), \( f''(x) \) changes from positive to negative: inflection point.
- From \( x = 4 \) to \( x = 5 \): \( f'(x) \) is increasing, so \( f''(x)>0 \). So at \( x = 4 \), \( f''(x) \) changes from negative to positive: inflection point.
- From \( x = 5 \) to \( x = 6 \): \( f'(x) \) is decreasing, so \( f''(x)<0 \). So at \( x = 5 \), \( f''(x) \) changes from positive to negative: inflection point.
- From \( x = 6 \) to \( x = 7 \): \( f'(x) \) is decreasing, so \( f''(x)<0 \) (wait, no, from \( x = 5 \) to \( x = 7 \), \( f'(x) \) is decreasing from \( x = 5 \) to \( x = 7 \)? Wait, at \( x = 5 \), \( f'(x) = 1 \), at \( x = 6 \), \( f'(x)=0 \), at \( x = 7 \), \( f'(x)= - 1 \). So from \( x = 5 \) to \( x = 7 \), \( f'(x) \) is decreasing, so \( f''(x)<0 \). Then from \( x = 7 \) to \( x = 8 \), \( f'(x) \) is increasing, so \( f''(x)>0 \). So at \( x = 7 \), \( f''(x) \) changes from negative to positive: inflection point.
- But this gives 6 points, but the answer options don't have 6. Wait, I must have made a mistake. Wait, the key is that a point of inflection for \( f(x) \) is where \( f''(x) \) changes sign, but also, the function \( f(x) \) must be twice - differentiable there (which it is, since \( f'(x) \) is differentiable as its graph is smooth). Wait, but the answer options are Two, Three, Four, Five, Six. Wait, maybe the local maxima and minima of \( f'(x) \) that are in the interval \([0,8]\) and where \( f''(x) \) changes sign. Wait, let's look at the graph again. Let's count the number of times \( f'(x) \) changes from increasing to decreasing or decreasing to increasing:
- At \( x = 1 \): \( f'(x) \) changes from increasing to decreasing (local max) → \( f''(x) \) changes from + to - → inflection point.
- At \( x = 2 \): \( f'(x) \) changes from decreasing to increasing (local min) → \( f''(x) \) changes from - to + → inflection point.
- At \( x = 3 \): \( f'(x) \) changes from increasing to decreasing (local max) → \( f''(x) \) changes from + to - → inflection point.
- At \( x = 4 \): \( f'(x) \) changes from decreasing to increasing (local min) → \( f''(x) \) changes from - to + → inflection point.
- At \( x = 5 \): \( f'(x) \) changes from increasing to decreasing (local max) → \( f''(x) \) changes from + to - → inflection point.
- At \( x = 7 \): \( f'(x) \) changes from decreasing to increasing (local min) → \( f''(x) \) changes from - to + → inflection point.
- But this is 6, but the answer options have E as Six? Wait, no, the option E is Six. Wait, but maybe I misread the graph. Wait, the graph of \( f'(x) \): Let's see the coordinates. At \( x = 1 \), peak at 3; \( x = 2 \), valley at 1; \( x = 3 \), peak at 2; \( x = 4 \), valley at 0; \( x = 5 \), peak at 1; \( x = 6 \), 0; \( x = 7 \), valley at - 1; \( x = 8 \), - 1? No, the graph: Let's see the local maxima of \( f'(x) \) are at \( x = 1 \), \( x = 3 \), \( x = 5 \) (3 points) and local minima at \( x = 2 \), \( x = 4 \), \( x = 7 \) (3 points). So that's 6 points. But the option E is Six. Wait, but maybe the question is about the number of inflection points where \( f(x) \) has a change in concavity, and all these 6 points are inflection points. But the option E is Six. Wait, but let's check the answer options again. The options are A:Two, B:Three, C:Four, D:Five, E:Six. So the correct answer should be E? Wait, no, maybe I made a mistake. Wait, no, let's recall: The graph of \( f'(x) \): Let's count the number of local extrema (maxima and minima) of \( f'(x) \) in \([0,8]\). The local maxima are at \( x = 1 \), \( x = 3 \), \( x = 5 \) (3 points) and local minima at \( x = 2 \), \( x = 4 \), \( x = 7 \) (3 points). So that's 6 points, so the number of inflection points of \( f(x) \) is 6. But wait, the option E is Six. So the answer is E? Wait, no, maybe I misread the graph. Wait, the graph of \( f'(x) \): Let's see, from \( x = 0 \) to \( x = 1 \), \( f'(x) \) is increasing; \( x = 1 \) to \( x = 2 \), decreasing; \( x = 2 \) to \( x = 3 \), increasing; \( x = 3 \) to \( x = 4 \), decreasing; \( x = 4 \) to \( x = 5 \), increasing; \( x = 5 \) to \( x = 7 \), decreasing; \( x = 7 \) to \( x = 8 \), increasing. So the number of times \( f'(x) \) changes from increasing to decreasing or decreasing to increasing is at \( x = 1 \) (inc to dec), \( x = 2 \) (dec to inc), \( x = 3 \) (inc to dec), \( x = 4 \) (dec to inc), \( x = 5 \) (inc to dec), \( x = 7 \) (dec to inc). That's 6 times, so 6 inflection points. So the answer is E.

Step3: Match with Options

The number of inflection points is 6, which corresponds to option E.

Answer:

E. Six