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The graph of the first derivative π prime of a function π is shown.
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What are the π₯-coordinates of the inflection points of π?
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Weβre given a graph of the first derivative π prime of a function π.
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We need to use this to determine the π₯-coordinates of all of the inflection points of our function π.
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To start, letβs recall what we mean by the π₯-coordinates of the inflection points of a function π.
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The inflection points for our function π will be the points where our function changes concavity and where our function is continuous.
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So the π₯-coordinates of these inflection points will be the points where π changes concavity at π₯ and where π is continuous at π₯.
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So there are two things which need to be true for our inflection points.
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Letβs start with π being continuous at π₯.
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In this question, weβre not told much about our function π of π₯.
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In fact, weβre not told anything about the continuity of π of π₯.
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All weβre given is a curve π¦ is equal to π prime of π₯.
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But we need to remember if a function is differentiable at a point, then it must in fact be continuous at this point.
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And we can see from our curve π of π₯ is differentiable for all values of π₯ greater than or equal to zero and less than or equal to nine.
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So π of π₯ is continuous on this interval as well.
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Itβs continuous on the closed interval from zero to nine.
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So in fact, in this case, we donβt need to worry about this stipulation for our inflection points.
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Our function is going to be continuous for all of our values of π₯.
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So in this case, we can just remove this condition altogether and worry entirely about the points where our curve changes concavity.
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To do this, weβre going to need to recall what we mean by the concavity of the curve π¦ is equal to π of π₯.
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We recall we say the curve π¦ is equal to π of π₯ is concave upward on an interval if on that interval all of its tangent lines lie below the curve.
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Similarly, we say the curve π¦ is equal to π of π₯ is concave downward on an interval if on that interval all of its tangent lines lie above the curve.
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So our inflection points for the curve π¦ is equal to π of π₯ will be where our curve switches from concave upward to concave downward or from concave downward to concave upward.
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But we can only use this directly if weβre given the curve π¦ is equal to π of π₯.
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But weβre not given this in this question.
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In this question, weβre instead given the curve π¦ is equal to π prime of π₯.
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So instead, we now need to think, what does the concavity of the curve π¦ is equal to π of π₯ mean for our curve π¦ is equal to π prime of π₯?
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To answer this, we need to recall what π prime of π₯ means.
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π prime of π₯ gives us the slope of the tangent line at π₯.
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So to determine what happens to π¦ is equal to π prime of π₯ when the concavity is changing, we need to determine what happens to the slope of our curve on intervals where our curve is concave upward and on intervals where our curve is concave downward.
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Letβs start with the slope of our tangent lines on intervals where our curve is concave upward.
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In our sketch, we can see that our tangent line start with negative slope.
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Eventually, their slope goes to zero.
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And then, after this, they go positive.
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In fact, we can recall that this is always true.
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On an interval where π¦ is equal to π of π₯ is concave upward, the slope of our tangent lines will be increasing.
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But we can then ask the question, what does this mean for the curve π¦ is equal to π prime of π₯?
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Remember, π prime of π₯ measures the slope of our tangent line.
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So if the slope of our tangent line is increasing, that means our function π prime of π₯ must also be increasing.
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So to find the intervals where π¦ is equal to π of π₯ is concave upward, we can instead find the intervals where the curve π¦ is equal to π prime of π₯ is increasing.
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And we can do something very similar to find the intervals where our curve π¦ is equal to π of π₯ is concave downward.
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On these intervals, the slope of our tangent lines will be decreasing.
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And since π prime of π₯ measures the slope of our tangent lines, this means on these intervals π¦ is equal to π prime of π₯ will be decreasing.
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So now all we need to do is look at the curve π¦ is equal to π prime of π₯ and determine where itβs increasing and where itβs decreasing.
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Letβs start from π₯ is equal to zero.
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We can see from π₯ is equal to zero all the way up to π₯ is equal to one, our curve is moving upward.
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In other words, π¦ is equal to π prime of π₯ is increasing.
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However, when π₯ is equal to one, we can see our curve turns.
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Itβs now moving downward.
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This means itβs decreasing.
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So in this case, when π₯ is equal to one, π prime of π₯ switches from being an increasing function to a decreasing function.
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And remember, when π prime of π₯ switches from being increasing to decreasing, this means π of π₯ is switching from being concave upward to concave downward.
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And if our function π is switching concavity at this value of π₯ and is continuous at this point, then itβs a point of inflection.
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So there must be a point of inflection for our function π when π₯ is equal to one.
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If we continue on in this way, we can find more points of inflections.
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We can see π prime of π₯ decreases all the way down to π₯ is equal to two.
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And then we can see when π₯ is greater than two, our curve π¦ is equal to π prime of π₯ is moving upward.
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Itβs increasing.
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So once again, we can see when π₯ is equal to two, π prime of π₯ switches from being decreasing to increasing.
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And of course, we know when π prime of π₯ switches from being decreasing to increasing, π of π₯ is switching from being concave downward to concave upward.
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So π is switching concavity at π₯ is equal to two.
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And therefore, π of π₯ has a point of inflection when π₯ is equal to two.
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And we could keep finding all of these points graphically.
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For example, we would find another inflection point when π₯ is equal to three.
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But itβs also worth pointing out we know how to find all of these points for our continuous curve π¦ is equal to π prime of π₯.
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These will be where our tangent lines that π prime of π₯ is equal to zero or, in other words, π double prime of π₯ is equal to zero.
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But remember, this will only work if π prime of π₯ is continuous.
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And we know π of π₯ and π prime of π₯ donβt have to be continuous.
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So this will only work in certain situations.
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So itβs worth keeping in mind both of these methods.
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They can be useful in different situations.
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Therefore, we were able to find the π₯-coordinates of the inflection points of our function π by looking at the intervals where the curve π¦ is equal to π prime of π₯ is increasing and decreasing.
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We were able to show that π has inflection points at π₯ is equal to one, π₯ is equal to two, π₯ is equal to three, π₯ is equal to five, and π₯ is equal to seven.