3.4 Maxima and Minima
Where a function peaks and bottoms out, where it rises and falls, and which way it bends — all read from $f^\prime$ and $f^{\prime\prime}$.
Given a function, we often want its largest and smallest values. They matter for accurate graphs and for optimization: maximizing profit, minimizing the aluminium in a can, finding the maximum height of a rocket. Here we use derivatives to find them.
Absolute Extrema
Consider $f(x)=x^2+1$ over $(-\infty,\infty)$. As $x\to\pm\infty$, $f(x)\to\infty$, so there is no largest value. But $x^2+1\ge1$ for all $x$, with equality at $x=0$, so $f$ has a smallest value, $1$, at $x=0$.
📘 Definition — Absolute extremum
Let $f$ be defined over an interval $I$ and let $c\in I$. $f$ has an absolute maximum on $I$ at $c$ if $f(c)\ge f(x)$ for all $x\in I$, and an absolute minimum on $I$ at $c$ if $f(c)\le f(x)$ for all $x\in I$. Either one is an absolute extremum.
"Absolute" here has nothing to do with absolute value — an extremum may be positive, negative or zero. The extremum is the value $f(c)$; the number $c$ is where it occurs.
📐 Theorem 3.1 — Extreme Value Theorem
If $f$ is a continuous function over the closed, bounded interval $[a,b]$, then there is a point in $[a,b]$ at which $f$ has an absolute maximum over $[a,b]$ and there is a point in $[a,b]$ at which $f$ has an absolute minimum over $[a,b]$.
Local Extrema and Critical Points
Picture a graph shaped like two mountains with a valley between them. The absolute maximum is at the higher peak, but the lower peak is interesting too: its value is larger than all nearby values. That is a local maximum; the bottom of the valley is a local minimum.
📘 Definition — Local extremum
$f$ has a local maximum at $c$ if there is an open interval $I$ containing $c$, contained in the domain of $f$, with $f(c)\ge f(x)$ for all $x\in I$; a local minimum if $f(c)\le f(x)$ for all $x\in I$. An absolute extremum at an endpoint is called an endpoint extremum, not a local extremum.
📘 Definition — Critical point
Let $c$ be an interior point in the domain of $f$. We say $c$ is a critical point of $f$ if $f^\prime(c)=0$ or $f^\prime(c)$ is undefined.
📐 Theorem 3.2 — Fermat's Theorem
If $f$ has a local extremum at $c$ and $f$ is differentiable at $c$, then $f^\prime(c)=0$.
📐 Theorem 3.3 — Location of absolute extrema
Let $f$ be continuous over a closed, bounded interval $I$. The absolute maximum and absolute minimum of $f$ over $I$ must occur at endpoints of $I$ or at critical points of $f$ in $I$.
🧭 Problem-Solving Strategy — Locating absolute extrema over a closed interval
- Evaluate $f$ at the endpoints $x=a$ and $x=b$.
- Find all critical points of $f$ in $(a,b)$ and evaluate $f$ there.
- Compare all the values. The largest is the absolute maximum; the smallest is the absolute minimum.
✏️ Example 1 — Locating absolute extrema
Find the absolute maximum and minimum over the given interval and state where they occur.
- a $f(x)=-x^2+3x-2$ over $[1,3]$
- b $f(x)=x^2-3x^{2/3}$ over $[0,2]$
💡 Show solution▶
(a) Step 1 Endpoints: $f(1)=0$ and $f(3)=-2$.
(a) Step 2 $f^\prime(x)=-2x+3$ is defined everywhere and zero at $x=\tfrac32\in[1,3]$; $f\left(\tfrac32\right)=\tfrac14$.
(a) Step 3 Compare $0,\ \tfrac14,\ -2$: the absolute maximum is $\tfrac14$ at $x=\tfrac32$; the absolute minimum is $-2$ at $x=3$.
(b) Step 1 Endpoints: $f(0)=0$ and $f(2)=4-3(2)^{2/3}\approx-0.762$.
(b) Step 2 $f^\prime(x)=2x-\dfrac{2}{x^{1/3}}=\dfrac{2x^{4/3}-2}{x^{1/3}}$ for $x\neq0$. It is zero at $x=\pm1$ and undefined at $x=0$. Only $x=1$ is interior to $[0,2]$: $f(1)=-2$.
(b) Step 3 Compare $0,\ -2,\ -0.762$: absolute maximum $0$ at $x=0$; absolute minimum $-2$ at $x=1$.
Derivatives and the Shape of a Graph
A local extremum must occur at a critical point, but a critical point need not give an extremum: $f(x)=x^3$ has $f^\prime(0)=0$ but no extremum at $0$. The sign of $f^\prime$ tells us which is which. If $f^\prime>0$ on an interval, $f$ is increasing there; if $f^\prime<0$, it is decreasing. A continuous function has a local maximum at $c$ exactly when it switches from increasing to decreasing there, and a local minimum when it switches from decreasing to increasing.
📐 Theorem 3.4 — First Derivative Test
Suppose $f$ is continuous over an interval $I$ containing a critical point $c$, and differentiable over $I$ except possibly at $c$.
- If $f^\prime$ changes from positive to negative at $c$, then $f(c)$ is a local maximum.
- If $f^\prime$ changes from negative to positive at $c$, then $f(c)$ is a local minimum.
- If $f^\prime$ has the same sign on both sides of $c$, then $f(c)$ is neither.
🧭 Problem-Solving Strategy — Using the first derivative test
- Find all critical points and use them to split $I$ into subintervals.
- Pick a test point in each subinterval and find the sign of $f^\prime$ there; this tells you where $f$ increases and decreases.
- Apply the first derivative test at each critical point.
✏️ Example 2 — Using the first derivative test to find local extrema
Use the first derivative test to find the location of all local extrema for $f(x)=x^3-3x^2-9x-1$.
💡 Show solution▶
Step 1 $f^\prime(x)=3x^2-6x-9=3(x-3)(x+1)$, so the critical points are $x=-1$ and $x=3$. They split the line into $(-\infty,-1)$, $(-1,3)$ and $(3,\infty)$.
Step 2 Test points: $f^\prime(-2)=(+)(-)(-)>0$ — increasing; $f^\prime(0)=(+)(-)(+)<0$ — decreasing; $f^\prime(4)=(+)(+)(+)>0$ — increasing.
Step 3 $f^\prime$ goes $+\to-$ at $x=-1$: a local maximum, $f(-1)=4$. It goes $-\to+$ at $x=3$: a local minimum, $f(3)=-28$.
Concavity and Points of Inflection
If the slope of the tangent increases as $x$ increases, the graph curves upward: $f$ is concave up. If the slope decreases, $f$ is concave down.
📘 Definition — Concavity
Let $f$ be differentiable over an open interval $I$. If $f^\prime$ is increasing over $I$, $f$ is concave up over $I$. If $f^\prime$ is decreasing over $I$, $f$ is concave down over $I$.
📐 Theorem 3.5 — Test for Concavity
Let $f$ be twice differentiable over an interval $I$. (i) If $f^{\prime\prime}(x)>0$ for all $x\in I$, $f$ is concave up over $I$. (ii) If $f^{\prime\prime}(x)<0$ for all $x\in I$, $f$ is concave down over $I$.
📘 Definition — Inflection point
If $f$ is continuous at $a$ and $f$ changes concavity at $a$, the point $(a,f(a))$ is an inflection point of $f$. Concavity can only change where $f^{\prime\prime}=0$ or is undefined — but it need not change there.
✏️ Example 3 — Testing for concavity
For $f(x)=x^3-6x^2+9x+30$, determine all intervals where $f$ is concave up and concave down, and list all inflection points.
💡 Show solution▶
Step 1 $f^\prime(x)=3x^2-12x+9$ and $f^{\prime\prime}(x)=6x-12$, which is zero only at $x=2$.
Step 2 Test points: $f^{\prime\prime}(0)=-12<0$ — concave down on $(-\infty,2)$; $f^{\prime\prime}(3)=6>0$ — concave up on $(2,\infty)$.
Step 3 Concavity changes at $x=2$, so $(2,f(2))=(2,32)$ is an inflection point.
| sign of $f^\prime$ | $f$ is | sign of $f^{\prime\prime}$ | concavity |
|---|---|---|---|
| + | increasing | + | concave up |
| + | increasing | − | concave down |
| − | decreasing | + | concave up |
| − | decreasing | − | concave down |
The Second Derivative Test
📐 Theorem 3.6 — Second Derivative Test
Suppose $f^\prime(c)=0$ and $f^{\prime\prime}$ is continuous over an interval containing $c$. (a) If $f^{\prime\prime}(c)>0$, $f$ has a local minimum at $c$. (b) If $f^{\prime\prime}(c)<0$, $f$ has a local maximum at $c$. (c) If $f^{\prime\prime}(c)=0$, the test is inconclusive.
✏️ Example 4 — Using the second derivative test
Use the second derivative to find the location of all local extrema for $f(x)=x^5-5x^3$.
💡 Show solution▶
Step 1 $f^\prime(x)=5x^4-15x^2=5x^2\left(x^2-3\right)=0$ at $x=0,\ \pm\sqrt3$.
Step 2 $f^{\prime\prime}(x)=20x^3-30x=10x\left(2x^2-3\right)$: $f^{\prime\prime}(-\sqrt3)=-30\sqrt3<0$ — local maximum; $f^{\prime\prime}(\sqrt3)=30\sqrt3>0$ — local minimum; $f^{\prime\prime}(0)=0$ — inconclusive.
Step 3 At $x=0$ use the first derivative test: $f^\prime(-1)<0$ and $f^\prime(1)<0$, so $f$ decreases on both sides and has no extremum at $0$.
Exercises
I
Find the critical points in the domains of the following functions.
- 1 $y=4x^3-3x$
- 2 $y=4\sqrt x-x^2$
- 3 $y=\dfrac{1}{x-1}$
- 4 $y=\ln(x-2)$
- 5 $y=\sqrt{4-x^2}$
- 6 $y=x+\dfrac1x$
See the critical points
Curve colour: blue = increasing, red = decreasing · background: green = concave up, purple = concave down · ◯ critical point · ◆ inflection point
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1 $y^\prime=12x^2-3=0\Rightarrow x=\pm\tfrac12$.
2 $y^\prime=\dfrac{2}{\sqrt x}-2x=0\Rightarrow\sqrt x\cdot x=1\Rightarrow x=1$. ($y^\prime$ is undefined at $x=0$, but that is an endpoint of the domain $[0,\infty)$, not an interior point.) Critical point: $x=1$.
3 $y^\prime=-\dfrac{1}{(x-1)^2}$ is never zero, and $x=1$ is not in the domain. No critical points.
4 $y^\prime=\dfrac{1}{x-2}$ is never zero and is defined on the whole domain $x>2$. No critical points.
5 $y^\prime=\dfrac{-x}{\sqrt{4-x^2}}=0\Rightarrow x=0$. ($y^\prime$ is undefined at $x=\pm2$, the endpoints.) Critical point: $x=0$.
6 $y^\prime=1-\dfrac{1}{x^2}=0\Rightarrow x=\pm1$ ($x=0$ is not in the domain).
II
Find the local and/or absolute extrema for the functions over the specified domain.
- 1 $f(x)=x^2+3$ over $[-1,4]$
- 2 $y=x^2+\dfrac2x$ over $[1,4]$
- 3 $y=\sqrt{9-x}$ over $[1,9]$
- 4 $y=\dfrac{x}{1+x}$ over $[0,100]$
- 5 $y=\sqrt x-\sqrt{x^3}$ over $[0,4]$
Closed-interval method
Curve colour: blue = increasing, red = decreasing · background: green = concave up, purple = concave down · ◯ critical point · ◆ inflection point
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1 $f^\prime=2x=0$ at $x=0$. Values: $f(-1)=4$, $f(0)=3$, $f(4)=19$. Absolute (and local) minimum $3$ at $x=0$; absolute maximum $19$ at $x=4$.
2 $y^\prime=2x-\dfrac{2}{x^2}=0$ only at $x=1$, an endpoint. $y$ is increasing on $(1,4)$. Absolute minimum $3$ at $x=1$; absolute maximum $\tfrac{33}{2}=16.5$ at $x=4$. No local extrema inside.
3 $y^\prime=\dfrac{-1}{2\sqrt{9-x}}<0$: always decreasing. Absolute maximum $\sqrt8=2\sqrt2$ at $x=1$; absolute minimum $0$ at $x=9$.
4 $y^\prime=\dfrac{1}{(1+x)^2}>0$: always increasing. Absolute minimum $0$ at $x=0$; absolute maximum $\tfrac{100}{101}$ at $x=100$.
5 $y^\prime=\dfrac{1}{2\sqrt x}-\dfrac32\sqrt x=\dfrac{1-3x}{2\sqrt x}=0$ at $x=\tfrac13$. Values: $y(0)=0$, $y\left(\tfrac13\right)=\dfrac{2\sqrt3}{9}\approx0.385$, $y(4)=2-8=-6$. Local and absolute maximum $\dfrac{2\sqrt3}{9}$ at $x=\tfrac13$; absolute minimum $-6$ at $x=4$.
III
Find the absolute maximum and absolute minimum of $f(x)=x^2-4x+3$ over the interval $[1,4]$.
💡 Show solution▶
Step 1 Endpoints: $f(1)=0$, $f(4)=3$.
Step 2 $f^\prime=2x-4=0$ at $x=2$; $f(2)=-1$.
Step 3 Absolute maximum $3$ at $x=4$; absolute minimum $-1$ at $x=2$. (See the third option of Interactive 2.)
IV
IV. Use the first derivative test to locate all local extrema for $f(x)=-x^3+\tfrac32x^2+18x$.
V. For the same $f$, find all intervals where $f$ is concave up and concave down.
VI. The points $c=3,\,-2$ satisfy $f^\prime(c)=0$. Use the second derivative test to decide whether $f$ has a local maximum or minimum there.
V. For the same $f$, find all intervals where $f$ is concave up and concave down.
VI. The points $c=3,\,-2$ satisfy $f^\prime(c)=0$. Use the second derivative test to decide whether $f$ has a local maximum or minimum there.
One function, three tests
Curve colour: blue = increasing, red = decreasing · background: green = concave up, purple = concave down · ◯ critical point · ◆ inflection point
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IV $f^\prime(x)=-3x^2+3x+18=-3(x-3)(x+2)$. Signs: $f^\prime(-3)<0$, $f^\prime(0)>0$, $f^\prime(4)<0$. So $-\to+$ at $x=-2$: local minimum $f(-2)=-22$; $+\to-$ at $x=3$: local maximum $f(3)=\tfrac{81}{2}$.
V $f^{\prime\prime}(x)=-6x+3=0$ at $x=\tfrac12$. Concave up on $\left(-\infty,\tfrac12\right)$, concave down on $\left(\tfrac12,\infty\right)$; inflection point at $x=\tfrac12$.
VI $f^{\prime\prime}(3)=-15<0$: local maximum at $x=3$. $f^{\prime\prime}(-2)=15>0$: local minimum at $x=-2$. Both tests agree.
VII
Determine (i) intervals where $f$ is increasing or decreasing, (ii) local minima and maxima, (iii) intervals of concavity, and (iv) inflection points.
- 1 $f(x)=x^3-4x^2+x+2$
- 2 $f(x)=x^2-6x$
- 3 $f(x)=x^3-6x^2$
- 4 $f(x)=x^4-6x^3$
- 5 $f(x)=x+x^2-x^3$
- 6 $f(x)=x^2+x+1$
- 7 $f(x)=x^3+x^4$
Full shape analysis
Curve colour: blue = increasing, red = decreasing · background: green = concave up, purple = concave down · ◯ critical point · ◆ inflection point
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1 $f^\prime=3x^2-8x+1=0$ at $x=\dfrac{4\pm\sqrt{13}}{3}\approx0.131,\ 2.535$. (i) Increasing on $(-\infty,0.131)\cup(2.535,\infty)$, decreasing between. (ii) Local max at $x=\tfrac{4-\sqrt{13}}{3}$ ($f\approx2.065$), local min at $x=\tfrac{4+\sqrt{13}}{3}$ ($f\approx-4.879$). (iii) $f^{\prime\prime}=6x-8$: concave down on $\left(-\infty,\tfrac43\right)$, up on $\left(\tfrac43,\infty\right)$. (iv) Inflection at $\left(\tfrac43,-\tfrac{38}{27}\right)$.
2 $f^\prime=2x-6$: decreasing on $(-\infty,3)$, increasing on $(3,\infty)$; local min $f(3)=-9$. $f^{\prime\prime}=2>0$: concave up everywhere; no inflection points.
3 $f^\prime=3x^2-12x=3x(x-4)$: increasing on $(-\infty,0)\cup(4,\infty)$, decreasing on $(0,4)$; local max $f(0)=0$, local min $f(4)=-32$. $f^{\prime\prime}=6x-12$: concave down on $(-\infty,2)$, up on $(2,\infty)$; inflection at $(2,-16)$.
4 $f^\prime=4x^3-18x^2=2x^2(2x-9)$: decreasing on $\left(-\infty,\tfrac92\right)$, increasing on $\left(\tfrac92,\infty\right)$; local min $f\left(\tfrac92\right)=-\tfrac{2187}{16}$; no extremum at $x=0$ (sign does not change). $f^{\prime\prime}=12x^2-36x=12x(x-3)$: up on $(-\infty,0)\cup(3,\infty)$, down on $(0,3)$; inflection points $(0,0)$ and $(3,-81)$.
5 $f^\prime=1+2x-3x^2=-(3x+1)(x-1)$: decreasing on $\left(-\infty,-\tfrac13\right)\cup(1,\infty)$, increasing on $\left(-\tfrac13,1\right)$; local min $f\left(-\tfrac13\right)=-\tfrac{5}{27}$, local max $f(1)=1$. $f^{\prime\prime}=2-6x$: up on $\left(-\infty,\tfrac13\right)$, down on $\left(\tfrac13,\infty\right)$; inflection at $\left(\tfrac13,\tfrac{11}{27}\right)$.
6 $f^\prime=2x+1$: decreasing on $\left(-\infty,-\tfrac12\right)$, increasing after; local min $f\left(-\tfrac12\right)=\tfrac34$. $f^{\prime\prime}=2$: concave up everywhere; no inflection points.
7 $f^\prime=3x^2+4x^3=x^2(4x+3)$: decreasing on $\left(-\infty,-\tfrac34\right)$, increasing on $\left(-\tfrac34,\infty\right)$; local min $f\left(-\tfrac34\right)=-\tfrac{27}{256}$; no extremum at $0$. $f^{\prime\prime}=6x+12x^2=6x(2x+1)$: up on $\left(-\infty,-\tfrac12\right)\cup(0,\infty)$, down on $\left(-\tfrac12,0\right)$; inflection points $\left(-\tfrac12,-\tfrac1{16}\right)$ and $(0,0)$.
VIII
Find all critical numbers. Remember to include interior points of the domain where $f^\prime$ does not exist.
- 1 $f(x)=x^{2/3}(x-5)$
- 2 $f(x)=\left|x^2-4\right|$
- 3 $g(t)=t\sqrt{4-t}$
- 4 $f(x)=x+2\cos x$ on $(0,2\pi)$
💡 Show solutions▶
1 $f(x)=x^{5/3}-5x^{2/3}$, so $f^\prime(x)=\tfrac53x^{2/3}-\tfrac{10}{3}x^{-1/3}=\dfrac{5x-10}{3x^{1/3}}=\dfrac{5(x-2)}{3\sqrt[3]{x}}$. $f^\prime=0$ at $x=2$; $f^\prime$ is undefined at $x=0$, which is in the domain. Critical numbers: $x=0$ and $x=2$.
2 Away from $x=\pm2$ the function is $\pm(x^2-4)$, so $f^\prime(x)=\pm2x$, which is zero only at $x=0$. At $x=\pm2$ the graph has corners (at both $x=2$ and $x=-2$ the left-hand slope is $-4$ and the right-hand slope is $4$), so $f^\prime(\pm2)$ does not exist. Critical numbers: $x=-2,\,0,\,2$.
3 Domain $t\le4$. $g^\prime(t)=\sqrt{4-t}-\dfrac{t}{2\sqrt{4-t}}=\dfrac{2(4-t)-t}{2\sqrt{4-t}}=\dfrac{8-3t}{2\sqrt{4-t}}$. $g^\prime=0$ at $t=\tfrac83$. ($g^\prime$ is undefined at $t=4$, but that is an endpoint of the domain.) Critical number: $t=\tfrac83$.
4 $f^\prime(x)=1-2\sin x=0\Rightarrow\sin x=\tfrac12\Rightarrow x=\dfrac\pi6$ or $x=\dfrac{5\pi}{6}$ in $(0,2\pi)$. $f^\prime$ exists everywhere, so these are the only critical numbers.
IX
Use the closed-interval method: list the critical numbers inside the interval, evaluate $f$ there and at both endpoints, and compare.
- 1 $f(x)=x^3-12x+1$ over $[-3,5]$
- 2 $f(x)=x+2\cos x$ over $[0,\pi]$
- 3 $f(x)=x^2e^{-x}$ over $[-1,3]$
- 4 $f(x)=3x^{2/3}-2x$ over $[-1,8]$
- 5 $f(x)=x-2\ln x$ over $[1,e]$
Closed-interval method with trig, exponential and cusp functions
Curve colour: blue = increasing, red = decreasing · background: green = concave up, purple = concave down · ◯ critical point · ◆ inflection point
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1 $f^\prime=3x^2-12=3(x-2)(x+2)=0$ at $x=\pm2$, both in the interval. Values: $f(-3)=10$, $f(-2)=17$, $f(2)=-15$, $f(5)=66$. Absolute maximum $66$ at $x=5$; absolute minimum $-15$ at $x=2$.
2 $f^\prime=1-2\sin x=0$ at $x=\tfrac\pi6,\tfrac{5\pi}6$. Values: $f(0)=2$, $f\left(\tfrac\pi6\right)=\tfrac\pi6+\sqrt3\approx2.256$, $f\left(\tfrac{5\pi}6\right)=\tfrac{5\pi}6-\sqrt3\approx0.886$, $f(\pi)=\pi-2\approx1.142$. Absolute maximum $\tfrac\pi6+\sqrt3$ at $x=\tfrac\pi6$; absolute minimum $\tfrac{5\pi}6-\sqrt3$ at $x=\tfrac{5\pi}6$.
3 $f^\prime=2xe^{-x}-x^2e^{-x}=x(2-x)e^{-x}=0$ at $x=0,2$ (since $e^{-x}\ne0$). Values: $f(-1)=e\approx2.718$, $f(0)=0$, $f(2)=4e^{-2}\approx0.541$, $f(3)=9e^{-3}\approx0.448$. Absolute maximum $e$ at $x=-1$; absolute minimum $0$ at $x=0$.
4 $f^\prime=2x^{-1/3}-2=\dfrac{2\left(1-\sqrt[3]{x}\right)}{\sqrt[3]{x}}$: zero at $x=1$, undefined at $x=0$. Values: $f(-1)=3(1)+2=5$, $f(0)=0$, $f(1)=1$, $f(8)=3(4)-16=-4$. Absolute maximum $5$ at $x=-1$; absolute minimum $-4$ at $x=8$.
5 $f^\prime=1-\dfrac2x=0$ at $x=2$, which lies in $[1,e]$ since $e\approx2.718$. Values: $f(1)=1$, $f(2)=2-2\ln2\approx0.614$, $f(e)=e-2\approx0.718$. Absolute maximum $1$ at $x=1$; absolute minimum $2-2\ln2$ at $x=2$.
X
Use the first derivative test. The last two items are an application and a conceptual check.
- 1 Find the local extrema of $f(x)=xe^{-x}$.
- 2 Find the local extrema of $f(x)=x-2\sin x$ on $(0,2\pi)$.
- 3 Find the local extrema of $f(x)=x^{4/3}-4x^{1/3}$.
- 4 Find the local extrema of $f(x)=x^2\ln x$, $x>0$.
- 5 The concentration of a drug in the bloodstream $t$ hours after an injection is $C(t)=\dfrac{20t}{t^2+4}$ mg/L for $0\le t\le6$. When is the concentration greatest, and what is it?
- 6 True or false? Explain. (a) If $f^\prime(c)=0$, then $f$ has a local extremum at $c$. (b) A function can have a local extremum at a point where $f^\prime$ does not exist.
First derivative test: watch the colour switch
Curve colour: blue = increasing, red = decreasing · background: green = concave up, purple = concave down · ◯ critical point · ◆ inflection point
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1 $f^\prime=e^{-x}-xe^{-x}=(1-x)e^{-x}$, zero only at $x=1$. Since $e^{-x}>0$, the sign of $f^\prime$ is the sign of $1-x$: $+$ for $x<1$, $-$ for $x>1$. So $+\to-$: local maximum $f(1)=\dfrac1e\approx0.368$. No local minimum.
2 $f^\prime=1-2\cos x=0\Rightarrow\cos x=\tfrac12\Rightarrow x=\tfrac\pi3,\tfrac{5\pi}3$. Test points: $f^\prime\left(\tfrac\pi6\right)=1-\sqrt3<0$, $f^\prime(\pi)=3>0$, $f^\prime\left(\tfrac{11\pi}6\right)=1-\sqrt3<0$. So local minimum $f\left(\tfrac\pi3\right)=\tfrac\pi3-\sqrt3\approx-0.685$ and local maximum $f\left(\tfrac{5\pi}3\right)=\tfrac{5\pi}3+\sqrt3\approx6.968$.
3 $f^\prime=\tfrac43x^{1/3}-\tfrac43x^{-2/3}=\dfrac{4(x-1)}{3x^{2/3}}$: zero at $x=1$, undefined at $x=0$. The denominator $3x^{2/3}$ is positive for $x\ne0$, so the sign is that of $x-1$: $-$ on $(-\infty,0)$, $-$ on $(0,1)$, $+$ on $(1,\infty)$. No sign change at $0$ (vertical tangent, no extremum); $-\to+$ at $1$: local minimum $f(1)=1-4=-3$.
4 $f^\prime=2x\ln x+x=x(2\ln x+1)$. For $x>0$ this is zero when $\ln x=-\tfrac12$, i.e. $x=e^{-1/2}=\dfrac1{\sqrt e}\approx0.607$. For $x<e^{-1/2}$, $2\ln x+1<0$ so $f^\prime<0$; for $x>e^{-1/2}$, $f^\prime>0$. Local minimum $f\left(e^{-1/2}\right)=e^{-1}\left(-\tfrac12\right)=-\dfrac{1}{2e}\approx-0.184$.
5 Quotient rule: $C^\prime(t)=\dfrac{20(t^2+4)-20t(2t)}{(t^2+4)^2}=\dfrac{20(4-t^2)}{(t^2+4)^2}$, zero at $t=2$ in $[0,6]$. $C^\prime>0$ on $(0,2)$ and $C^\prime<0$ on $(2,6)$, so $t=2$ is a local maximum. Compare with the endpoints: $C(0)=0$, $C(2)=\dfrac{40}{8}=5$, $C(6)=\dfrac{120}{40}=3$. The greatest concentration is $5$ mg/L, reached $2$ hours after the injection.
6 (a) False. $f(x)=x^3$ has $f^\prime(0)=0$, but $f^\prime=3x^2\ge0$ on both sides of $0$, so $f$ is increasing through $0$ and there is no extremum. A critical number is only a candidate. (b) True. $f(x)=|x|$ has an absolute (and local) minimum at $x=0$, yet $f^\prime(0)$ does not exist because of the corner. This is why points where $f^\prime$ is undefined must be included among the critical numbers.
🔬 Research–Teaching NexusResearch-oriented · optional, about 30 minutes
Research question. On which day of the year is Nizwa hottest, according to a model and according to data?
Why it matters. Climate scientists fit periodic models to temperature data to study seasons and warming trends; energy companies use them to predict peak electricity demand for air conditioning.
What to do.
- Use this sample model for daily maximum temperature: $T(d)=34+9\sin\left(\frac{2\pi(d-105)}{365}\right)$ °C, $d$ = day of the year.
- Find $T'(d)$ and solve $T'(d)=0$ to find the hottest and coolest days.
- Find real monthly average maximum temperatures for Nizwa from a weather website or the Oman Meteorology site.
- Plot the real data with the model in Desmos. Does the hottest month agree?
Tools: Desmos, weather website
Hand in: A half-page note: the hottest day from the model, the hottest month from the data, and one reason they may differ.
Going further. Anywhere: fit the same model to your own city's temperatures.
Instructor note: The model's maximum is $43$ °C at $d\approx196$ (mid-July); students should notice real data are not a perfect sine curve.
🧪 Self-check quiz
0/5 correct
Q1. A critical point of $f$ is an interior point $c$ where:
Q2. The Extreme Value Theorem guarantees absolute extrema for a function that is:
Q3. If $f^\prime(c)=0$ and $f^{\prime\prime}(c)<0$, then at $c$:
Q4. For $f(x)=x^3$, the point $x=0$ is:
Q5. $f(x)=x^3-6x^2+9x+30$ has an inflection point at: