5.1 Critical Valuesap Calculus

A local maximum point on a function is apoint \$(x,y)\$ on the graph of the function whose \$y\$ coordinate islarger than all other \$y\$ coordinates on the graph at points 'closeto' \$(x,y)\$. More precisely, \$(x,f(x))\$ is a local maximum if thereis an interval \$(a,b)\$ with \$a< x< b\$ and \$f(x)ge f(z)\$ for every \$z\$in \$(a,b)\$. Similarly, \$(x,y)\$ is a local minimum pointif it has locally the smallest \$y\$ coordinate. Againbeing more precise: \$(x,f(x))\$ is a local minimum if thereis an interval \$(a,b)\$ with \$a< x< b\$ and \$f(x)le f(z)\$ for every \$z\$in \$(a,b)\$. A local extremumis either a local minimum or a local maximum.

Local maximum and minimum points are quite distinctive on the graph ofa function, and are therefore useful in understanding the shape of thegraph. In many applied problems we want to find the largest orsmallest value that a function achieves (for example, we might wantto find the minimum cost at which some task can be performed) and soidentifying maximum and minimum points will be useful for appliedproblems as well. Some examples of local maximum and minimum pointsare shown in figure 5.1.1.

The critical point is correctly classified as a relative maximum, and the student’s justification is correct. In part (c) the student earned the antiderivative point with the correct expression presented on the right side of the second line. The limit expression point was earned with the expression on the left side of the second line.

Figure 5.1.1. Some local maximum points (\$A\$) and minimum points (\$B\$).

If \$(x,f(x))\$ is a point where \$f(x)\$ reaches a local maximum or minimum,and if the derivative of \$f\$ exists at \$x\$, then the graph has atangent line and the tangent line must be horizontal. This isimportant enough to state as a theorem, though we will not prove it.

Theorem 5.1.1 (Fermat's Theorem) If \$f(x)\$ has a local extremum at \$x=a\$ and\$f\$ is differentiable at \$a\$, then \$f'(a)=0\$.

Thus, the onlypoints at which a function can have a local maximum or minimum arepoints at which the derivative is zero, as in the left hand graph infigure 5.1.1,or the derivative is undefined, as in the right hand graph. Any valueof \$x\$ for which \$f'(x)\$ is zero or undefined is called acritical value for \$f\$.When looking for local maximum and minimum points, you are likely tomake two sorts of mistakes: You may forget that a maximum or minimumcan occur where the derivative does not exist, and so forget to checkwhether the derivative exists everywhere. You might also assume thatany place that the derivative is zero is a local maximum or minimumpoint, but this is not true. A portion of the graph of \$ds f(x)=x^3\$ isshown in figure 5.1.2. The derivative of \$f\$ is\$f'(x)=3x^2\$, and \$f'(0)=0\$, but there is neither a maximum norminimum at \$(0,0)\$.

Figure 5.1.2. No maximum or minimum even though the derivative is zero.

Since the derivative is zero or undefined at both local maximum andlocal minimum points, we need a way to determine which, if either,actually occurs. The mostelementary approach, but one that is often tedious or difficult, is totest directly whether the \$y\$ coordinates 'near' the potentialmaximum or minimum are above or below the \$y\$ coordinate at the pointof interest. Of course, there are too many points 'near' the pointto test, but a little thought shows we need only test two provided weknow that \$f\$ is continuous (recall that this means that the graph of\$f\$ has no jumps or gaps).

AP Calculus Critical Points and Extreme Value Theorem Notes Name Date Period ©v R2J0x1q8y GKkuNtHaf hSKo`fTtdwHaNrie kLWLrCe.X k MAl`l` rciMgkhytGsQ JrMeysqeFrEvteCdP.-1-Determine all of the critical points for the function. 1) g (x) = 6x5 + 33x4 - 30x3 + 100 x = -5, 0, 3 5 2) f (t) = 3 t2 (2t - 1) t =. 5.1 Extreme Values Find the extreme values and where th local max of -l when x = -2 absolute min of -3 when x = 0 -o absolute max of I when x = 2 local min of-2 when x = 4 occur. Local max of 3 when x = -2 local min of 2 when x = O PRACTICE local min of -1 when x local max of 1 when x local min of 0 when x -5 absolute max of 2 when x = 2. Calculus Examples. Popular Problems. Find the Critical Points f(x)=x-5x^(1/5). The critical points of a function are where the value of makes the.

Suppose, for example, that we have identified three points at which\$f'\$ is zero or nonexistent: \$ds (x_1,y_1)\$, \$ds (x_2,y_2)\$, \$ds (x_3,y_3)\$,and \$ds x_1< x_2< x_3\$ (see figure 5.1.3). Suppose that we compute the value of \$f(a)\$ for \$ds x_1< a< x_2\$, andthat \$ds f(a)< f(x_2)\$. What can we say about the graph between \$a\$ and\$ds x_2\$? Could there be a point \$ds (b,f(b))\$, \$ds a< b< x_2\$ with\$ds f(b)>f(x_2)\$? No: if there were, the graph would go up from\$(a,f(a))\$ to \$(b,f(b))\$ then down to \$ds (x_2,f(x_2))\$ and somewhere inbetween would have a local maximum point. (This is not obvious; it isa result of the Extreme Value Theorem, theorem 6.1.2.)But at that local maximumpoint the derivative of \$f\$ would be zero or nonexistent, yet wealready know that the derivative is zero or nonexistent only at \$ds x_1\$,\$ds x_2\$, and \$ds x_3\$. The upshot is that one computation tells us that\$ds (x_2,f(x_2))\$ has the largest \$y\$ coordinate of any point on thegraph near \$ds x_2\$ and to the left of \$ds x_2\$. We can perform the sametest on the right. If we find that on both sides of \$ds x_2\$ the valuesare smaller, then there must be a local maximum at \$ds (x_2,f(x_2))\$; ifwe find that on both sides of \$ds x_2\$ the values are larger, then theremust be a local minimum at \$ds (x_2,f(x_2))\$; if we find one of each,then there is neither a local maximum or minimum at \$ds x_2\$.

Figure 5.1.3. Testing for a maximum or minimum.

It is not always easy to compute the value of a function at aparticular point. The task is made easier by the availability ofcalculators and computers, but they have their own drawbacks—they donot always allow us to distinguish between values that are very closetogether. Nevertheless, because this method is conceptually simple andsometimes easy to perform, you should always consider it.

Example 5.1.2 Find all local maximum and minimum points for the function \$ds f(x)=x^3-x\$. The derivative is \$ds f'(x)=3x^2-1\$. This is definedeverywhere and is zero at \$ds x=pm sqrt{3}/3\$. Looking first at\$ds x=sqrt{3}/3\$, we see that \$ds f(sqrt{3}/3)=-2sqrt{3}/9\$. Now we testtwo points on either side of \$ds x=sqrt{3}/3\$, making sure that neither is farther away thanthe nearest critical value; since \$ds sqrt{3}< 3\$, \$ds sqrt{3}/3< 1\$ andwe can use \$x=0\$ and \$x=1\$. Since\$ds f(0)=0>-2sqrt{3}/9\$and \$ds f(1)=0>-2sqrt{3}/9\$, there must be a local minimum at \$ds x=sqrt{3}/3\$. For \$ds x=-sqrt{3}/3\$, we see that\$ds f(-sqrt{3}/3)=2sqrt{3}/9\$. This time we can use \$x=0\$ and \$x=-1\$,and we find that \$ds f(-1)=f(0)=0< 2sqrt{3}/9\$, so there must be a localmaximum at \$ds x=-sqrt{3}/3\$.

Of course this example is made very simple by our choice of points totest, namely \$x=-1\$, \$0\$, \$1\$. We could have used other values, say\$-5/4\$, \$1/3\$, and \$3/4\$, but this would have made the calculationsconsiderably more tedious.

Example 5.1.3 Find all local maximum and minimum points for \$f(x)=sin x+cos x\$. The derivative is \$f'(x)=cos x-sin x\$. This isalways defined and is zero whenever \$cos x=sin x\$. Recalling thatthe \$cos x\$ and \$sin x\$ are the \$x\$ and \$y\$ coordinates of points ona unit circle, we see that \$cos x=sin x\$ when \$x\$ is \$pi/4\$, \$pi/4pmpi\$, \$pi/4pm2pi\$, \$pi/4pm3pi\$, etc. Since both sineand cosine have a period of \$2pi\$, we need only determine the statusof \$x=pi/4\$ and \$x=5pi/4\$. We can use \$0\$ and \$pi/2\$ to test thecritical value \$x= pi/4\$. We find that \$ds f(pi/4)=sqrt{2}\$, \$ds f(0)=1< sqrt{2}\$ and \$ds f(pi/2)=1\$,so there is a local maximum when \$x=pi/4\$ and also when\$x=pi/4pm2pi\$, \$pi/4pm4pi\$, etc. We can summarize this moreneatly by saying that there are local maxima at \$pi/4pm 2kpi\$ forevery integer \$k\$.

We use \$pi\$ and \$2pi\$ to test the critical value \$x=5pi/4\$. Therelevant values are \$ds f(5pi/4)=-sqrt2\$, \$ds f(pi)=-1>-sqrt2\$,\$ds f(2pi)=1>-sqrt2\$, so there is a local minimum at \$x=5pi/4\$,\$5pi/4pm2pi\$, \$5pi/4pm4pi\$, etc. More succinctly, there arelocal minima at \$5pi/4pm 2kpi\$ forevery integer \$k\$.

Exercises 5.1

In problems 1–12, find all local maximum and minimumpoints \$(x,y)\$ by the method of this section.

Ex 5.1.1\$ds y=x^2-x\$ (answer)

Ex 5.1.2\$ds y=2+3x-x^3\$ (answer)

Ex 5.1.4\$ds y=x^4-2x^2+3\$ (answer)

Ex 5.1.7\$ds y=3x^2-(1/x^2)\$ (answer)

Ex 5.1.9\$ds f(x) =cases{ x-1 & \$x < 2\$ crx^2 & \$xgeq 2\$cr}\$(answer)

Ex 5.1.10\$ds f(x) =cases{x-3 & \$x < 3\$ crx^3 & \$3leq x leq 5\$cr1/x &\$x>5\$cr}\$(answer)

Ex 5.1.11\$ds f(x) = x^2 - 98x + 4\$(answer)

Ex 5.1.12\$ds f(x) =cases{ -2 & \$x = 0\$ cr1/x^2 &\$x neq 0\$cr}\$(answer)

Ex 5.1.13For any real number \$x\$ there is a unique integer \$n\$ such that \$n leq x < n +1\$, and the greatest integer function is defined as \$dslfloor xrfloor = n\$. Where are the critical values of the greatest integer function? Which are local maxima and which are local minima?

Ex 5.1.14Explain why the function \$f(x) =1/x\$ has no localmaxima or minima.

Ex 5.1.15How many critical points can a quadratic polynomial function have?(answer)

Ex 5.1.16Show that a cubic polynomial can have at most two criticalpoints. Give examples to show that a cubic polynomial can have zero,one, or two critical points.

Ex 5.1.17Explore the family of functions \$ds f(x) = x^3 + cx +1\$ where \$c\$ is a constant. How many and what types of local extremes are there? Your answer should depend on the value of \$c\$, that is, different values of \$c\$ will give different answers.

Ex 5.1.18We generalize the preceding two questions. Let \$n\$ be apositive integer and let \$f\$ be a polynomial of degree \$n\$. How manycritical points can \$f\$ have? (Hint: Recall the Fundamental Theorem of Algebra, which says that a polynomial of degree \$n\$ has at most \$n\$ roots.)

A local maximum point on a function is apoint \$(x,y)\$ on the graph of the function whose \$y\$ coordinate islarger than all other \$y\$ coordinates on the graph at points 'closeto' \$(x,y)\$. More precisely, \$(x,f(x))\$ is a local maximum if thereis an interval \$(a,b)\$ with \$a< x< b\$ and \$f(x)ge f(z)\$ for every \$z\$in \$(a,b)\$. Similarly, \$(x,y)\$ is a local minimum pointif it has locally the smallest \$y\$ coordinate. Againbeing more precise: \$(x,f(x))\$ is a local minimum if thereis an interval \$(a,b)\$ with \$a< x< b\$ and \$f(x)le f(z)\$ for every \$z\$in \$(a,b)\$. A local extremumis either a local minimum or a local maximum.

Local maximum and minimum points are quite distinctive on the graph ofa function, and are therefore useful in understanding the shape of thegraph. In many applied problems we want to find the largest orsmallest value that a function achieves (for example, we might wantto find the minimum cost at which some task can be performed) and soidentifying maximum and minimum points will be useful for appliedproblems as well. Some examples of local maximum and minimum pointsare shown in figure 5.1.1.

Figure 5.1.1. Some local maximum points (\$A\$) and minimum points (\$B\$).

If \$(x,f(x))\$ is a point where \$f(x)\$ reaches a local maximum or minimum,and if the derivative of \$f\$ exists at \$x\$, then the graph has atangent line and the tangent line must be horizontal. This isimportant enough to state as a theorem, though we will not prove it.

Theorem 5.1.1 (Fermat's Theorem) If \$f(x)\$ has a local extremum at \$x=a\$ and\$f\$ is differentiable at \$a\$, then \$f'(a)=0\$.

Thus, the onlypoints at which a function can have a local maximum or minimum arepoints at which the derivative is zero, as in the left hand graph infigure 5.1.1,or the derivative is undefined, as in the right hand graph. Any valueof \$x\$ for which \$f'(x)\$ is zero or undefined is called acritical value for \$f\$.When looking for local maximum and minimum points, you are likely tomake two sorts of mistakes: You may forget that a maximum or minimumcan occur where the derivative does not exist, and so forget to checkwhether the derivative exists everywhere. You might also assume thatany place that the derivative is zero is a local maximum or minimumpoint, but this is not true. A portion of the graph of \$ds f(x)=x^3\$ isshown in figure 5.1.2. The derivative of \$f\$ is\$f'(x)=3x^2\$, and \$f'(0)=0\$, but there is neither a maximum norminimum at \$(0,0)\$.

Figure 5.1.2. No maximum or minimum even though the derivative is zero.

Since the derivative is zero or undefined at both local maximum andlocal minimum points, we need a way to determine which, if either,actually occurs. The mostelementary approach, but one that is often tedious or difficult, is totest directly whether the \$y\$ coordinates 'near' the potentialmaximum or minimum are above or below the \$y\$ coordinate at the pointof interest. Of course, there are too many points 'near' the pointto test, but a little thought shows we need only test two provided weknow that \$f\$ is continuous (recall that this means that the graph of\$f\$ has no jumps or gaps).

Suppose, for example, that we have identified three points at which\$f'\$ is zero or nonexistent: \$ds (x_1,y_1)\$, \$ds (x_2,y_2)\$, \$ds (x_3,y_3)\$,and \$ds x_1< x_2< x_3\$ (see figure 5.1.3). Suppose that we compute the value of \$f(a)\$ for \$ds x_1< a< x_2\$, andthat \$ds f(a)< f(x_2)\$. What can we say about the graph between \$a\$ and\$ds x_2\$? Could there be a point \$ds (b,f(b))\$, \$ds a< b< x_2\$ with\$ds f(b)>f(x_2)\$? No: if there were, the graph would go up from\$(a,f(a))\$ to \$(b,f(b))\$ then down to \$ds (x_2,f(x_2))\$ and somewhere inbetween would have a local maximum point. (This is not obvious; it isa result of the Extreme Value Theorem, theorem 6.1.2.)But at that local maximumpoint the derivative of \$f\$ would be zero or nonexistent, yet wealready know that the derivative is zero or nonexistent only at \$ds x_1\$,\$ds x_2\$, and \$ds x_3\$. The upshot is that one computation tells us that\$ds (x_2,f(x_2))\$ has the largest \$y\$ coordinate of any point on thegraph near \$ds x_2\$ and to the left of \$ds x_2\$. We can perform the sametest on the right. If we find that on both sides of \$ds x_2\$ the valuesare smaller, then there must be a local maximum at \$ds (x_2,f(x_2))\$; ifwe find that on both sides of \$ds x_2\$ the values are larger, then theremust be a local minimum at \$ds (x_2,f(x_2))\$; if we find one of each,then there is neither a local maximum or minimum at \$ds x_2\$.

Figure 5.1.3. Testing for a maximum or minimum.

It is not always easy to compute the value of a function at aparticular point. The task is made easier by the availability ofcalculators and computers, but they have their own drawbacks—they donot always allow us to distinguish between values that are very closetogether. Nevertheless, because this method is conceptually simple andsometimes easy to perform, you should always consider it.

Example 5.1.2 Find all local maximum and minimum points for the function \$ds f(x)=x^3-x\$. The derivative is \$ds f'(x)=3x^2-1\$. This is definedeverywhere and is zero at \$ds x=pm sqrt{3}/3\$. Looking first at\$ds x=sqrt{3}/3\$, we see that \$ds f(sqrt{3}/3)=-2sqrt{3}/9\$. Now we testtwo points on either side of \$ds x=sqrt{3}/3\$, making sure that neither is farther away thanthe nearest critical value; since \$ds sqrt{3}< 3\$, \$ds sqrt{3}/3< 1\$ andwe can use \$x=0\$ and \$x=1\$. Since\$ds f(0)=0>-2sqrt{3}/9\$and \$ds f(1)=0>-2sqrt{3}/9\$, there must be a local minimum at \$ds x=sqrt{3}/3\$. For \$ds x=-sqrt{3}/3\$, we see that\$ds f(-sqrt{3}/3)=2sqrt{3}/9\$. This time we can use \$x=0\$ and \$x=-1\$,and we find that \$ds f(-1)=f(0)=0< 2sqrt{3}/9\$, so there must be a localmaximum at \$ds x=-sqrt{3}/3\$.

Of course this example is made very simple by our choice of points totest, namely \$x=-1\$, \$0\$, \$1\$. We could have used other values, say\$-5/4\$, \$1/3\$, and \$3/4\$, but this would have made the calculationsconsiderably more tedious.

Example 5.1.3 Find all local maximum and minimum points for \$f(x)=sin x+cos x\$. The derivative is \$f'(x)=cos x-sin x\$. This isalways defined and is zero whenever \$cos x=sin x\$. Recalling thatthe \$cos x\$ and \$sin x\$ are the \$x\$ and \$y\$ coordinates of points ona unit circle, we see that \$cos x=sin x\$ when \$x\$ is \$pi/4\$, \$pi/4pmpi\$, \$pi/4pm2pi\$, \$pi/4pm3pi\$, etc. Since both sineand cosine have a period of \$2pi\$, we need only determine the statusof \$x=pi/4\$ and \$x=5pi/4\$. We can use \$0\$ and \$pi/2\$ to test thecritical value \$x= pi/4\$. We find that \$ds f(pi/4)=sqrt{2}\$, \$ds f(0)=1< sqrt{2}\$ and \$ds f(pi/2)=1\$,so there is a local maximum when \$x=pi/4\$ and also when\$x=pi/4pm2pi\$, \$pi/4pm4pi\$, etc. We can summarize this moreneatly by saying that there are local maxima at \$pi/4pm 2kpi\$ forevery integer \$k\$.

We use \$pi\$ and \$2pi\$ to test the critical value \$x=5pi/4\$. Therelevant values are \$ds f(5pi/4)=-sqrt2\$, \$ds f(pi)=-1>-sqrt2\$,\$ds f(2pi)=1>-sqrt2\$, so there is a local minimum at \$x=5pi/4\$,\$5pi/4pm2pi\$, \$5pi/4pm4pi\$, etc. More succinctly, there arelocal minima at \$5pi/4pm 2kpi\$ forevery integer \$k\$.

Exercises 5.1

In problems 1–12, find all local maximum and minimumpoints \$(x,y)\$ by the method of this section.

Ex 5.1.1\$ds y=x^2-x\$ (answer)

Ex 5.1.2\$ds y=2+3x-x^3\$ (answer)

Ex 5.1.4\$ds y=x^4-2x^2+3\$ (answer)

Ex 5.1.7\$ds y=3x^2-(1/x^2)\$ (answer)

Ex 5.1.9\$ds f(x) =cases{ x-1 & \$x < 2\$ crx^2 & \$xgeq 2\$cr}\$(answer)

Ex 5.1.10\$ds f(x) =cases{x-3 & \$x < 3\$ crx^3 & \$3leq x leq 5\$cr1/x &\$x>5\$cr}\$(answer)

5.1 Critical Valuesap Calculus 2nd Edition

Ex 5.1.11\$ds f(x) = x^2 - 98x + 4\$(answer)

5.1 Critical Valuesap Calculus Solver

Ex 5.1.12\$ds f(x) =cases{ -2 & \$x = 0\$ cr1/x^2 &\$x neq 0\$cr}\$(answer)

Ex 5.1.13For any real number \$x\$ there is a unique integer \$n\$ such that \$n leq x < n +1\$, and the greatest integer function is defined as \$dslfloor xrfloor = n\$. Where are the critical values of the greatest integer function? Which are local maxima and which are local minima?

Ex 5.1.14Explain why the function \$f(x) =1/x\$ has no localmaxima or minima.

5.1 Critical Valuesap Calculus Transcendentals

Ex 5.1.15How many critical points can a quadratic polynomial function have?(answer)

Ex 5.1.16Show that a cubic polynomial can have at most two criticalpoints. Give examples to show that a cubic polynomial can have zero,one, or two critical points.

5.1 Critical Valuesap Calculus 14th Edition

Ex 5.1.17Explore the family of functions \$ds f(x) = x^3 + cx +1\$ where \$c\$ is a constant. How many and what types of local extremes are there? Your answer should depend on the value of \$c\$, that is, different values of \$c\$ will give different answers.

Ex 5.1.18We generalize the preceding two questions. Let \$n\$ be apositive integer and let \$f\$ be a polynomial of degree \$n\$. How manycritical points can \$f\$ have? (Hint: Recall the Fundamental Theorem of Algebra, which says that a polynomial of degree \$n\$ has at most \$n\$ roots.)