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Find, if they exist, the local maximum and/or minimum values of the function π of π₯ equals nine π to the nine π₯ plus nine π to the negative nine π₯.
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Also specify what type of value they are.
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We have a function, and we want to find its local maxima and minima.
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The local maxima and minima will occur at critical numbers of the function; that is, where the derivative of the function is zero or does not exist.
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So letβs differentiate.
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We differentiate term by term.
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What is the derivative of nine π to the nine π₯?
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Itβs nine times the derivative of π to the nine π₯, which is nine π to the nine π₯, and so we get 81π to the nine π₯.
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Here weβve used a special case of the chain rule, the derivative of a function of a number π times π₯ is that number π times the derivative of π evaluated at ππ₯.
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In our application, π was nine and the function π was the exponential function π to the π₯.
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Using the same rule, we can differentiate the second term nine π to the negative nine π₯.
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Thatβs nine times negative nine π to the negative nine π₯, because the derivative of π to the negative nine π₯ is negative nine π to the negative nine π₯.
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Simplifying, we get 81π to nine π₯ minus 81π to the negative nine π₯.
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Remember, weβre looking for local maxima and minima of our function, which occur at the critical numbers where the derivative is zero or doesnβt exist.
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We can see that our derivative exists for all real values of π₯.
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Thereβs no value of π₯ for which this is undefined.
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And so the local maxima and minima, if they exist, occur when the derivative is zero.
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So letβs solve this equation.
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We can divide through by 81, add π to the negative nine π₯ to both sides.
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We divide through by π to the nine π₯ using laws of exponents to write π to the negative nine π₯ divided by π to the nine π₯ as π to the negative 18π₯.
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We take the natural logarithm on both sides and see that π₯ must be zero.
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So we have a critical number of zero.
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And this is the only place where the function can have a local maximum or a local minimum.
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But we donβt know which happens at zero; it could be a local maximum or a local minimum.
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And in fact it could be neither!
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So how do we tell?
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One way to do this, though not the only way, is to use the second derivative test.
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Suppose that π prime of π, the derivative of π at π, is zero and π double prime, the second derivative of π, is continuous near π.
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In other words, itβs continuous in an open interval containing π.
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If the value of the second derivative of π at π is greater than zero, then π has a local minimum at π.
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But if the second derivative of π is less than zero at π, then π has a local maximum at π.
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If, however, the second derivative of π at π is zero, then we canβt say anything.
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They could be a local maximum or a local minimum or neither.
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Letβs clear some room so we can apply our test.
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The second derivative test involves the second derivative of π, hence the name.
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So we need to differentiate π prime to get π double prime, the second derivative of π.
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And this is a very similar process to differentiating π.
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The derivative of 81π to the nine π₯ is 81 times nine π to the nine π₯.
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And from this, we subtract the derivative of 81π to the negative nine π₯, which is 81 times negative nine times π to the negative nine π₯.
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And the two minus signs make a plus, so we get 729π to the nine π₯ plus 729π to the negative nine π₯.
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Okay, now that we have the second derivative of π, letβs apply the second derivative test.
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Suppose that π prime of π is zero.
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Well thatβs true when π is the critical number zero and π double prime is continuous near π.
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Is our second derivative π double prime continuous near zero?
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Yes, itβs continuous, not just around the critical number zero, but everywhere.
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The conditions for the second derivative test are satisfied then.
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The test itself relies on the sign of π double prime at π.
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Now what is the sign of π double prime at zero?
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We substitute zero for π₯ and see that we can simplify the exponents: nine times zero is zero as is negative nine times zero; thatβs also zero.
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So π double prime of zero is 729π to the zero plus 729π to the zero.
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And anything to the power of zero is one.
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So we have 729 times one plus 729 times one.
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And thatβs just 729 plus 729, which is 1458.
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The precise value of π double prime of zero isnβt important.
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What is important is the sign of this value.
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And in our case, we see that itβs positive.
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Looking across at the second derivative test, we see that if π double prime of π is greater than zero, and weβve just seen that it is, then π has a local minimum at π.
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Now what is this local minimum value?
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Well itβs π of zero.
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So we substitute zero for π₯ in the expression for π of π₯ that we have.
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Itβs nine π to the nine times zero plus nine π to the negative nine times zero.
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And as we saw before, both π to the nine times zero and π to the negative nine times zero are one.
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And so the value of our function at zero is nine plus nine equals 18.
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Whatβs our final answer: π of zero equals 18, and this is a local minimum value of our function π of π₯ equals nine π to the nine π₯ plus nine π to the negative nine π₯.
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This is the only local minimum of our function, and there are no local maxima.
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Letβs recap!
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We differentiated π to find the critical number zero, which is the only place at which a local maximum or minimum can occur.
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And then we used the second derivative test to test this critical value, showing that there was a local minimum there.
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The second derivative test depends on the sign of the second derivative at π.
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You can understand the second derivative test intuitively by considering the second derivative π double prime as being the rate of change of the slope function π prime.
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At a local minimum of a function π, the slope π prime is zero because the slope of the tangent to the curve at this point is zero.
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To the left of this point, the slope is negative.
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And to the right, it is positive.
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The slope, and therefore the slope function π prime, is increasing as we pass π.
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So the rate of change of the slope function π prime is positive, and the rate of change of the slope function π prime is just the second derivative: π double prime.
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So π double prime is positive at π.
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Similarly, near a local maximum, the slope function π prime decreases as π₯ increases, starting positive but decreasing to zero as we reach the maximum and then decreasing further to become a negative as we pass this local maximum point.
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Hence at a local maximum, π double prime, which is the rate of change of the slope function π prime, is negative.