Critical andInflection Points

1 Finding and Clas-sifying Critical PointsA critical point is a point on thegraph where the tangent slopeis horizontal, (0) or vertical,(∞). or not defined like theminimum of y = |x|.

0

1

2

3

4

-4 -2 2 4x

Some people call criticalpoints stationary points or ex-trema but we will always callthem critical points. Firstwhat’s so special about criticalpoints? Why are they criti-cal? They are critical becausethey could be maximum orminimum points.If the tangent slope at a

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point is horizontal, that is,y0 = 0, then the curve has oneof four shapes near that criticalpoint.

You can see that the firsttwo situations are a maximumand minimum respectively.The bottom two situations are

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neither, they are just placeson the graph with a horizontaltangent slope.If the tangent slope at the

critical point is vertical, thatis, y0 = ∞, then there arefour possible shapes near thecritical point.

The top two are called4

cusps. The first one has acritical point that is a maxi-mum and has a vertical tangentslope. The second is a min-imum and also has a verticaltangent slope. The bottom twoare just critical points wherethe tangent slope is vertical.They are neither maximum norminimum.Ok, let’s do the first part of

this activity, find the criticalpoints. All we have to dois take the derivative of thefunction in question and see

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what x0s make the tangentslope 0 or ∞. That is, set thederivative equal to zero, orsee where the derivative isundefined. By undefined weusually mean the denominatorof a fraction is zero. Let’s dosome examples to flesh thisout.

ExampleQ? Find the critical

points of the function y =2x3 − 9x2 + 12x + 6.A. Take the derivative

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of the function to gety0 = 6x2 − 18x+ 12

Now, the derivative or tangentslope function is a polynomialso y0 is never going to be ∞.Set y0 = 0 and solve.

0 = 6x2 − 18x + 12= 6(x− 1)(x− 2)

Therefore at the points wherex = 1 and x = 2 the tangentslope is horizontal. This givesus the x-values of the twocritical points. What are they-values? We get the y-valuesby plugging the x-values into

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the original equation, y =2x3− 9x2 + 12x+ 6. Thus the twocritical points are (1, 11) and(2, 10).

Now notice the graph of thefunction from Example andspot the critical points.

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8

10

12

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0 0.5 1 1.5 2 2.5 3x

ExampleQ? Find the critical

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points of the functiony =

2

3x32 − 2x12

A. Take the derivativey0 = x

12 − x−12

and then factor out the “x tothe lowest power”, in our case,x−12.

y0 = x−12

hx12 − 1

i=

hx12 − 1

ix12

We can see that if x = 0, thenthe denominator of the abovefraction would be zero, butx = 0 is a defined point on the

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graph. Plug x = 0 into theoriginal equation and see thatthe graph goes through theorigin. We conclude that at thecritical point where x = 0 wehave a vertical tangent slope.To check for horizontal tangentslopes we set y0 = 0 and solve.Note that for a fraction tobe zero, the top part must be

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zero.

y0 =

hx12 − 1

ix12

0 =

hx12 − 1

ix12

0 = x12 − 1

1 = x12

x = 1

Therefore (1,−43) is a criticalpoint where the tangent slopeis horizontal.When we look at the graph

of the function from Examplewe can see the critical pointsbut we also notice that there

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are no points where the x-values are negative. Why?

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-0.5

0

0.5

1

-1 1 2 3 4x

Because the domain of thefunction is [0,∞). Look backto the beginning of the termto refresh your recollection ofrange and domains.Ok, now we can find critical

points. The next step is toclassify them. We need toknow whether a critical point

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is a• local maximum• local minimum• or neitherTo do this we will draw a

slope line. This is an importantillustrative tool. After theslope line is drawn we willuse the First Derivative Testto classify the critical points.There is a Second DerivativeTest that we will learn later

ExampleQ? Classify the critical

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points of the function y =2x3 − 9x2 + 12x + 6

1.1 First DerivativeTestAfter we draw the slope linewe can use these patterns toclassify the critical points• / - \ indicates a maximum

• \ - / indicates a minimum

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• \ | / indicates a cusp that’s aminimum

• / | \ indicates a cusp that’s amaximum

• / - / indicates a critical pointthat is neither a maximumnor a minimum

• \ - \ indicates a critical pointthat is neither a maximum

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nor a minimum

• / | / indicates a critical pointthat is neither a maximumnor a minimum

• \ | \ indicates a critical pointthat is neither a maximumnor a minimum

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Steps for drawing a SlopeLine

1. Find all the critical points ofthe function.

2. Draw a number line with thecritical points on it and theirslopes represented above, ei-ther as a horizontal or verticalline.

3. Pick test points ( x-values )in between the critical pointsand plug them into the firstderivative to tell if the tangentslope is positive or negative inthat interval.

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4. Add those slopes to the num-ber line to represent the tan-gent slopes in the intervals

5. Use the First derivative test toclassify the critical points.

This concept of a slope linecan answer many questionsconcerning a particular func-tion. We have already used itto classify critical point withthe First Derivative Test. Wecan also use the slope lineto find intervals of increaseor decrease, that is, intervals

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where the graph is increasingor decreasing respectively.

ExampleQ? Find the intervals

of increase and decrease of thefunction y = 2x3 − 9x2 + 12x + 6from Example .Another benefit of the slope

line is that we get a rough ideaof the shape of the graph. Theslope line shown in Exampleimplies the curve increasesuntil x = 1 where it reachesa peak and drops down untilx = 2 where it bottoms out

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and increases continually afterx = 2. This implication isconfirmed by the actual graphshown after Example .

Local versus Global

We use the terms local andglobal to describe maximumor minimum points. All max-imum or minimum points arelocal maximums or minimumsrespectively. They are thehighest (or lowest) spot on thegraph in their immediate vicin-ity. The highest point on the

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whole graph is the global max-imum. Likewise, the lowestpoint on the whole graph is theglobal minimum.

2 Concavity andPoints of InflectionThe first derivative is the tan-gent slope formula, the secondderivative is the concavity for-mula. Concavity is the bendin the curve. Positive concav-ity or concave up occurs wheny00 > 0 and negative concav-

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ity or concave down occurswhen y00 < 0. There can beany combination of slope andconcavity.• Positive concavity with posi-tive slope, y00 > 0, y0 > 0

• Positive concavity with neg-ative slope, y00 > 0, y0 < 0

• Negative concavity withpositive slope, y00 < 0, y0 > 0

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• Negative concavity withnegative slope, y00 < 0, y0 < 0

A good way to rememberthe difference between pos-itive concavity and negativeconcavity is this cheesy child’sdrawing.

Anyway, just like when wedrew the slope line, we drawa concavity line. The steps

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to make it are similar to theslope line. Get the secondderivative of the function andfind the x-values that make y00zero or undefined and markthose points on your concavityline. These points are calledpossible points of inflection.Use test points on either sidesof these possible points ofinflection to determine theconcavity of the intervals.Wait you say. What’s an

inflection point?

Definition 1 An inflection point is24

a point on the graph where theconcavity changes from concaveup to concave down or from con-cave down to concave up.

So a possible point of in-flection still has to prove itselfto be a point of inflection byhaving different concavities oneither side of it.ExampleQ? Find the intervals

of positive concavity and theintervals of negative concavity

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of the functiony =

x5

20− x

4

12+ 7x

and any inflection points.A.We can see the concavity in

the graph of the function fromExample .

-100

-50

0

50

100

-4 -2 2 4x

Oh, I promised we wouldlearn the Second Deriva-tive Test, eh. Here it is, as

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promised.A point on a curve with a

horizontal tangent slope andnegative concavity must be amaximum.

Likewise a point with ahorizontal tangent slope andpositive concavity must be aminimum.

This thinking forms thebasis for the second derivative

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test. First find the criticalpoints, then test them with the:

2.1 Second DerivativeTestTo test a critical point x =a, y0(a) = 0 to classify it as amaximum or a minimum:• If y00(a) < 0 then x = a is amaximum point

• If y00(a) > 0 then x = a is aminimum point

That’s it. This test is usedfrequently in economics. So

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now we have two ways toclassify critical points. Mypersonal favorite is the FirstDerivative Test. It’s muchbetter. You only have to takeone derivative. Notice theSecond Derivative Test is onlygood for critical points wherethe slope is horizontal andthere is no conclusion to bemade when y00(a) = 0. Thenwhat? You have to do theFirst Derivative Test anyway.Also, you can get so muchmore information from the

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slope line than you can withthe Second Derivative Test.I’ll show you an example ofthe inferior Second DerivativeTest just for completeness.

ExampleQ? Find and classify

the critical points of the curvey = x4 − 2x2 + 3.

Homework Section 2.10#10, 13, 14 - 18, 21, 22Submit 10, 16, 22Section 3.4 #25 - 29Submit 26

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Section 3.5 #49, 50Section 3.6 #3 - 12Submit 4, 6, 10

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