WEBVTT
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Determine the intervals on which the function π of π₯ equals three π₯ cubed minus nine π₯ squared minus four is increasing and on which it is decreasing.
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By definition, a function is increasing on an interval πΌ if π of π₯ one is less than π of π₯ two whenever π₯ one is less than π₯ two for π₯ one and π₯ two in the interval πΌ.
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And it is decreasing on πΌ if π of π₯ one is greater than π of π₯ two whenever π₯ one is less than π₯ two for π₯ one and π₯ two in πΌ.
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Thatβs the formal definition.
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But really the best way to think about increasing and decreasing functions is using their graphs.
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The graph of an increasing function looks like this: π¦ increases as π₯ increases.
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Theyβre not always at the same rate.
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And the graph of the function which is decreasing on the interval looks like this.
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Now letβs turn to the problem we have.
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We have to determine the intervals on which our function is increasing and on which it is decreasing.
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We could try to graph this function using a graphing calculator or graphing software.
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And in fact, weβll do that at the end of the video.
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But the way weβre going to solve this question is using a test.
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If the derivative of π, π prime of π₯, is greater than zero on an interval, πΌ then π is increasing on πΌ.
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And if the derivative of π, π prime of π₯, is less than zero on an interval πΌ, then π is decreasing on that interval πΌ.
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And this test makes sense given the graphical interpretation above.
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A tangent to the increasing function always has positive slope, so π prime is positive.
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But the tangents to the decreasing function have negative slope, so π prime is negative on this interval for this decreasing function.
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Letβs clear away the definition and graphical interpretation so we have room to apply our test.
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The test involves π prime, which is the derivative of our function.
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So letβs find π prime.
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We differentiate the polynomial term by term, using the fact that the derivative of ππ₯ to the π with respect to π₯ is π times π times π₯ to the π minus one.
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The derivative of three π₯ to the power of three comes from multiplying the exponent and coefficient to get nine and then reducing the exponent by one to get π₯ squared.
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Similarly, the derivative of nine π₯ squared is 18π₯, and the derivative of four is zero.
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So π prime of π₯ is nine π₯ squared minus 18π₯.
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π is increasing on intervals for which this derivative is greater than zero and decreasing on intervals for which the derivative is less than zero.
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So we need to find the sign of π prime.
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The way we do this is to factor π prime, writing it as nine π₯ times π₯ minus two.
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And from this factored form, we can easily see that π prime is zero at π₯ equals zero and π₯ equals two.
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These two numbers are called the critical numbers of the function π.
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For π prime to change sign from positive to negative or negative to positive, it must pass through zero.
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And so we know that π prime must have the same sign on the interval π₯ is less than zero.
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If π prime changed sign below zero, because itβs continuous, there will be a value of π₯ below zero at which π prime was zero.
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And weβve seen that π prime is only zero at π₯ equals zero and π₯ equals two.
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Similarly, π prime canβt change sign between zero and two.
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Similarly, π prime canβt change sign when π₯ is greater than two.
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If π prime does change sign anywhere, itβs got to be at zero or two.
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We want to find the sign of π prime on these intervals.
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And we do that by considering the signs of its factors.
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So along the top of the table, we have the two factors of π prime: nine π₯ and π₯ minus two and π prime itself.
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Letβs start filling the table in.
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What is the sign of nine π₯ when π₯ is less than zero?
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Nine π₯ is negative when π₯ is less than zero.
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You can see this by multiplying both sides of the inequality, which defines this interval, by nine.
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What can we say about nine π₯ when π₯ is between zero and two?
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In a similar way, we see that nine π₯ is between zero and 18; in particular, its positive.
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And finally for the third interval, we see that is nine π₯ is greater than 18 and so is positive.
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We do the same thing for the factor π₯ minus two.
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In the first region, it is less than negative two and so negative.
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In the second region, it is between negative two and zero, and so itβs negative here too.
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And finally in the third region, itβs positive.
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We found the sign of the factors of π prime.
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Now we just need to find the sign of π prime itself in each of these regions.
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In the first region, π₯ is less than zero, both factors are negative.
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And so π prime is negative times negative, which is a positive.
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Similarly, in the second region, π prime is a positive times a negative quantity, which makes it a negative quantity.
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And finally, in the third region, π prime is a positive times a positive, and so it is a positive quantity.
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Referring back to the increasing decreasing test, if π prime is positive on πΌ then π is increasing on πΌ; and if π prime is negative on πΌ, then π is decreasing on πΌ.
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So π is increasing on these two intervals and decreasing on this one.
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The only thing left to do is to write down our final answer: π₯ is increasing on the intervals where π₯ is less than zero and where π₯ is greater than two and is decreasing on the interval where π₯ is between zero and two.
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Letβs write these using interval notation.
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π₯ is less than zero represents the open interval between negative infinity and zero.
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π₯ is greater than two is the open interval from two to infinity.
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And zero is less than π₯ is less than two represents the open interval from zero to two.
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So hereβs our answer: π is increasing on the open interval from negative infinity to zero, and it is also increasing on the open interval from two to infinity; π is, however, decreasing on the open interval from zero to two.
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We solved this problem using the increasing/decreasing test, which uses the sign of π prime to tell where π is increasing and where it is decreasing.