6 Problems
Existence, and where to look
The problems follow the sections of the theory. Those that are checked at once get new numbers each time you press New variant: repeat them until the method runs by itself. Those with a solution are for paper. Write the whole solution before you open it, then compare line by line. Last come problems in the form of the exam, with points and a marking guide.
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A sign change forces a root
Read the theory: 1. A sign change forces a root →For each function, decide whether Fact 6.1 guarantees a root in the given interval, and say why. Then say whether there is a root there.
(a) $f(x)=x^{3}-4x+1$ on $[0,1]$
(b) $g(x)=\dfrac{1}{x-2}$ on $[1,3]$
(c) $h(x)=x^{2}-2x$ on $[-1,3]$
Hint
Fact 6.1 needs two things: continuity on the whole closed interval, and values of opposite signs at the two ends. If either fails, the Fact says nothing — and then you have to look for yourself.
(a) The Fact applies. $f$ is a polynomial, hence continuous on $[0,1],$ and $f(0)=1>0$ while $f(1)=1-4+1=-2<0.$ The signs differ, so there is a root in $(0,1).$
(b) The Fact does not apply. $g(1)=-1$ and $g(3)=1$ have opposite signs, but $g$ is not defined at $x=2,$ so it is not continuous on $[1,3].$ The Fact is silent. Looking for ourselves: a fraction with numerator $1$ is never zero, so there is no root.
(c) The Fact does not apply either. $h$ is continuous, but $h(-1)=1+2=3$ and $h(3)=9-6=3$ have the same sign. Again the Fact is silent — and this time there are roots: $h(x)=x(x-2)$ is zero at $x=0$ and at $x=2,$ both in $[-1,3].$
The moral. When a hypothesis fails, the theorem says nothing. In (b) there is no root and in (c) there are two: silence is not a “no”.
Answer
(a) Guaranteed: continuous, $f(0)=1,$ $f(1)=-2.$ There is a root in $(0,1).$
(b) Not guaranteed ($g$ is not continuous at $2$), and there is no root.
(c) Not guaranteed (no sign change), but there are two roots, $0$ and $2.$
How did it go?
Stuck? Read 1. A sign change forces a root →
Newton's method
Read the theory: 2. Newton's method →Newton himself tried his method on the equation $$x^{3}-2x-5=0\,.$$
(a) Take two steps of Newton's method from $x_{0}=2.$
(b) Show with Fact 6.1 that there is a root between $2.0945$ and $2.0946,$ and conclude what the root is to three decimals.
Hint
In (a), $f'(x)=3x^{2}-2.$ Keep five or six decimals. In (b), compute $f$ at the two numbers: you only need the signs. A sign change is a certificate that no amount of iterating can give on its own.
(a) The first step. $f(2)=8-4-5=-1$ and $f'(2)=12-2=10,$ so $$x_{1}=2-\frac{-1}{10}=2.1\,.$$
(a) The second step. $f(2.1)=9.261-4.2-5=0.061$ and $f'(2.1)=13.23-2=11.23,$ so $$\begin{aligned} x_{2}&=2.1-\frac{0.061}{11.23}\\ &\approx2.1-0.005432\\ &=2.094568\,. \end{aligned}$$
(b) The sign change. $f$ is a polynomial, hence continuous, and $$\begin{aligned} f(2.0945)&\approx-0.00057<0\,,\\ f(2.0946)&\approx0.00054>0\,. \end{aligned}$$ By Fact 6.1 there is a root $c$ with $2.0945<c<2.0946.$
(b) Three decimals. Every number strictly between $2.0945$ and $2.0946$ rounds to $2.095.$ So the root is $2.095$ to three decimals. The iteration found the digits; the sign change proves them.
Answer
(a) $x_{1}=2.1$ and $x_{2}\approx2.09457.$
(b) $f(2.0945)<0<f(2.0946),$ so there is a root in between: to three decimals it is $2.095.$
How did it go?
Stuck? Read 2. Newton's method →
Let $f(x)=\sqrt[3]{x}=x^{1/3},$ whose only root is $x=0.$ Show that Newton's method from any starting point $x_{0}\ne0$ gives $$x_{n}=(-2)^{n}x_{0}\,,$$ so that the iterates run away from the root. Explain in words what the tangents do.
Hint
Compute $f(x)/f'(x)$ for $x\ne0$ first. It simplifies to a multiple of $x.$ Then think about how steep the graph of $\sqrt[3]{x}$ is near $0$ and far from it.
The step. For $x\ne0$ we have $f'(x)=\tfrac13x^{-2/3},$ so $$\begin{aligned} \frac{f(x)}{f'(x)}&=\frac{x^{1/3}}{\tfrac13x^{-2/3}}\\ &=3x^{1/3+2/3}=3x\,. \end{aligned}$$
The iteration. Hence $x_{n}=x_{n-1}-3x_{n-1}=-2x_{n-1}.$ Each step multiplies by $-2,$ so $x_{n}=(-2)^{n}x_{0}.$ The distance to the root doubles at every step, and the sign alternates: from $x_{0}=1$ the iterates are $1,-2,4,-8,\ldots$ (For $x<0$ the same computation holds, since $\sqrt[3]{x}$ is defined and differentiable there too.)
The tangents. The graph is very steep near $0$ and flatter further out. The tangent at $x_{0}$ is too flat to reach the axis near $0{:}$ it crosses at $-2x_{0},$ on the other side and twice as far away. There the tangent is flatter still. Nothing is wrong with $f'(x_{n-1})\ne0;$ the starting point is simply never sensible (Watch out 6.7).
Answer
$f(x)/f'(x)=3x,$ so $x_{n}=x_{n-1}-3x_{n-1}=-2x_{n-1}$ and $x_{n}=(-2)^{n}x_{0}.$ Each tangent overshoots the root to twice the distance on the other side.
How did it go?
Stuck? Read 2. Newton's method →
Existence again: the extreme value theorem
Read the theory: 3. Existence again: the extreme value theorem →For each function and interval, say whether Fact 6.9 guarantees a maximum and a minimum. Then find the maximum and the minimum value where they exist, and say why the others do not exist.
(a) $f(x)=x^{2}$ on $[-1,2]$
(b) $f(x)=x^{2}$ on $(-1,2)$
(c) $g(x)=\dfrac{1}{x}$ on $[1,\infty)$
Hint
Check the three hypotheses: continuous, closed, bounded. A sketch shows what happens near an end that is missing.
(a) Guaranteed. $x^{2}$ is continuous and $[-1,2]$ is closed and bounded. The candidates are $x=0$ (where $f'(x)=2x=0$) and the endpoints: $f(-1)=1,$ $f(0)=0,$ $f(2)=4.$ The maximum is $4,$ at $x=2,$ and the minimum is $0,$ at $x=0.$
(b) Not guaranteed: $(-1,2)$ is not closed. The minimum still exists: $x^{2}\ge0=f(0)$ for all $x,$ and $0$ is in the interval. There is no maximum: the values approach $4$ as $x$ approaches $2,$ but $2$ is not in the interval. Take any $x$ in $(-1,2).$ If $x\le1,$ then $f(x)\le1<f(1.5)=2.25.$ If $1<x<2,$ the point $y=\tfrac{x+2}{2}$ lies between $x$ and $2,$ so $y^{2}>x^{2}.$ Either way some point of the interval beats $x,$ so no point is a maximum point.
(c) Not guaranteed: $[1,\infty)$ is not bounded. The maximum exists: $g$ is decreasing, so $g(x)\le g(1)=1.$ There is no minimum: every value $1/x$ is positive, and $g(2x)=\tfrac{1}{2x}$ is smaller still, so no value is the smallest.
The moral. When Fact 6.9 does not apply, each of the two extremes may exist or not. The Fact is silent; it is the function that decides.
Answer
(a) Guaranteed; maximum $4$ at $x=2,$ minimum $0$ at $x=0.$
(b) Not guaranteed; minimum $0$ at $x=0,$ no maximum.
(c) Not guaranteed; maximum $1$ at $x=1,$ no minimum.
How did it go?
Stuck? Read 3. Existence again: the extreme value theorem →
Where to look
Read the theory: 4. Where to look →Three kinds of candidate
Read the theory: 5. Three kinds of candidate →Find the largest and the smallest value of $$f(x)=|x-1|+\frac{x}{2}$$ on the interval $[-2,3].$
Hint
Write $f$ without the absolute value on each side of $x=1.$ Is there any point where $f'(x)=0?$ Is there a point where $f'$ does not exist? Method 6.15 has three kinds of candidate.
The guarantee. $f$ is continuous and $[-2,3]$ is closed and bounded, so both extremes exist (Fact 6.9).
Each side of $x=1.$ For $x<1$ we have $|x-1|=1-x,$ so $f(x)=1-\tfrac{x}{2}$ and $f'(x)=-\tfrac12.$ For $x>1$ we have $|x-1|=x-1,$ so $f(x)=\tfrac{3x}{2}-1$ and $f'(x)=\tfrac32.$
The candidates. $f'$ is never $0.$ At $x=1$ the slope jumps from $-\tfrac12$ to $\tfrac32,$ so the graph has a corner and $f'(1)$ does not exist: a candidate of the third kind. With the endpoints the list is $-2,$ $1$ and $3{:}$ $$\begin{aligned} f(-2)&=3-1=2\,,\\ f(1)&=0+\tfrac12=\tfrac12\,,\\ f(3)&=2+\tfrac32=\tfrac72\,. \end{aligned}$$
Compare. The largest value is $\tfrac72,$ at the endpoint $x=3.$ The smallest is $\tfrac12,$ at the corner $x=1,$ where there is no derivative to set to zero.
Answer
Largest value $\tfrac72$ at $x=3;$ smallest value $\tfrac12$ at the corner $x=1.$
How did it go?
Stuck? Read 5. Three kinds of candidate →
When a candidate is the answer
Read the theory: 6. When a candidate is the answer →Let $f(x)=\ln x-x+1$ for $x>0.$
(a) Show that $f$ is concave on $(0,\infty),$ and find its maximum point.
(b) Conclude that $\ln x\le x-1$ for every $x>0,$ with equality only at $x=1.$
(c) What does (b) say about the graph of $\ln x$ and its tangent at $x=1?$
Hint
The interval has no endpoints, so look for Fact 6.20 rather than a list of candidates. For “equality only at $x=1$”, look at the sign of $f'$ on each side of $1.$
(a) Concave. $f'(x)=\dfrac1x-1$ and $f''(x)=-\dfrac{1}{x^{2}}<0$ for all $x>0,$ so $f$ is concave on $(0,\infty)$ (Fact 6.20(i)).
(a) The maximum point. $f'(x)=0$ gives $x=1,$ an interior point of $(0,\infty).$ By Fact 6.20(ii) it is a maximum point over the whole interval, and $f(1)=0-1+1=0.$
(b) The inequality. So $f(x)\le f(1)=0$ for every $x>0,$ that is $\ln x-x+1\le0,$ or $\ln x\le x-1.$
(b) Equality only at $1.$ $f'(x)=\dfrac{1-x}{x}$ is positive on $(0,1)$ and negative on $(1,\infty),$ so $f$ is strictly increasing up to $1$ and strictly decreasing after it (Fact 2.23). Hence $f(x)<f(1)=0$ for every $x\ne1.$
(c) The tangent. The tangent of $\ln x$ at $x=1$ is $y=\ln1+\tfrac11(x-1)=x-1.$ So (b) says that the graph of $\ln x$ lies below its tangent at $x=1,$ touching it only there — as it must, since $(\ln x)''=-1/x^{2}<0$ and the curve bends downwards (Fact 3.22).
Answer
(a) $f''(x)=-1/x^{2}<0;$ the maximum point is $x=1,$ with $f(1)=0.$
(b) $\ln x\le x-1,$ with equality only at $x=1.$
(c) The graph of $\ln x$ lies below its tangent $y=x-1$ at $x=1$ and touches it only there.
How did it go?
Stuck? Read 6. When a candidate is the answer →
Mixed, as on the exam
(a) Let $p>0$ and let $q$ be any number. Show that the equation $$x^{3}+px+q=0$$ has exactly one real solution.
(b) Give an example with $p<0$ where the equation has three real solutions. Which step of your argument in (a) fails?
Hint
Two tasks. At most one: the sign of the derivative. At least one: you need a number where the cubic is negative and one where it is positive. Try $M=|q|+1$ and $-M,$ and use $M^{3}\ge M$ for $M\ge1.$
(a) At most one. Put $f(x)=x^{3}+px+q.$ Then $f'(x)=3x^{2}+p>0$ for every $x,$ since $p>0.$ So $f$ is strictly increasing (Fact 2.23) and has at most one root.
(a) A positive value. Let $M=|q|+1,$ so $M\ge1$ and $M^{3}\ge M.$ Then, since $pM>0,$ $$\begin{aligned} f(M)&=M^{3}+pM+q\\ &>M+q\\ &=|q|+q+1\ge1\,. \end{aligned}$$
(a) A negative value. In the same way $$\begin{aligned} f(-M)&=-M^{3}-pM+q\\ &<-M+q\\ &=q-|q|-1\le-1\,. \end{aligned}$$
(a) At least one. $f$ is a polynomial, hence continuous on $[-M,M],$ and $f(-M)<0<f(M).$ By Fact 6.1 it has a root in $(-M,M).$ With the first step: exactly one.
(b) Three solutions. Take $p=-7,$ $q=6{:}$ $$\begin{aligned} &x^{3}-7x+6\\ &\quad=(x-1)(x-2)(x+3)\,, \end{aligned}$$ with the three roots $1,$ $2$ and $-3.$ The “at least one” half still holds for every $p$ (far enough out, the cube wins), but $f'(x)=3x^{2}-7$ is negative near $0,$ so $f$ is not increasing and the “at most one” step fails.
Answer
(a) $f'(x)=3x^{2}+p>0$ gives at most one root, and $f(-M)<0<f(M)$ with $M=|q|+1$ gives at least one.
(b) For example $x^{3}-7x+6=(x-1)(x-2)(x+3)$: the step “at most one” fails, because $f'$ changes sign.
How did it go?
Stuck? Read 1. A sign change forces a root →
A workshop can produce at most $10$ units a week. Its weekly profit from $Q$ units, for $0\le Q\le10,$ is $$\pi(Q)=30Q-Q^{2}-50\,.$$
(a) Find the output that maximises profit, and the profit.
(b) If the capacity limit were removed, so that every $Q\ge0$ were possible, what would the best output and the profit be? Justify that it is the best.
(c) About how much would one more unit of capacity be worth a week? Compare with the exact gain from $Q=10$ to $Q=11.$
Hint
In (a), Method 6.15: is the stationary point inside the interval? In (b) there is no right end; use Fact 6.20. In (c), the derivative is the profit of one more unit.
(a) The guarantee and the candidates. $\pi$ is a polynomial, hence continuous, and $[0,10]$ is closed and bounded. $\pi'(Q)=30-2Q$ is zero only at $Q=15,$ which is outside $[0,10].$ So the candidates are the endpoints: $$\begin{aligned} \pi(0)&=-50\,,\\ \pi(10)&=300-100-50=150\,. \end{aligned}$$ The best output is $Q=10,$ at full capacity, with profit $150.$
(b) Without the limit. On $[0,\infty)$ we have $\pi''(Q)=-2<0,$ so $\pi$ is concave, and $Q=15$ is an interior point with $\pi'(15)=0.$ By Fact 6.20(ii) it is the maximum point over the whole of $[0,\infty),$ with $\pi(15)=450-225-50=175.$
(c) One more unit. At $Q=10$ the marginal profit is $\pi'(10)=30-20=10{:}$ one more unit of capacity is worth about $10$ a week. Exactly, $\pi(11)=330-121-50=159,$ a gain of $9.$ The derivative overstates it a little because $\pi'$ falls as $Q$ grows.
Answer
(a) $Q=10$ (the capacity limit), with profit $150.$
(b) $Q=15,$ with profit $175;$ it is the global maximum since $\pi''=-2<0.$
(c) About $10$ a week ($\pi'(10)=10$); exactly $9.$
How did it go?
Stuck? Read 5. Three kinds of candidate →
As on the exam
3 problems · 23 pProblems in exam form, with points. Solve them on paper or in the box. Then open the solution one step at a time and mark yourself against the rubric.
A rational function with two parameters: conditions, sign table, extreme points
For real numbers $a$ and $b,$ let $$f(x)=\frac{x^{2}+ax+b}{x-2}\,,\qquad x\ne2\,.$$
- (a)3 p
Find $a$ and $b$ such that $f$ has a stationary point at $x=-1$ with $f(-1)=-1.$
- (b)4 p
With these values of $a$ and $b,$ make a sign table for $f'$ and give the intervals on which $f$ is increasing and on which it is decreasing. Find the largest value of $f$ on the interval $(-\infty,2)$ and the smallest value of $f$ on $(2,\infty),$ and the points where they are taken.
- (c)3 p
Does $f$ have a maximum point on its whole domain? A minimum point? Justify your answers.
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(a) The derivative. By the quotient rule, with the numerator's derivative $2x+a$ and the denominator's derivative $1,$ $f'(x)=\dfrac{N(x)}{(x-2)^{2}}$ with $$\begin{aligned} N(x)&=(2x+a)(x-2)\\ &\quad-(x^{2}+ax+b)\\ &=2x^{2}-4x+ax-2a\\ &\quad-x^{2}-ax-b\\ &=x^{2}-4x-2a-b\,. \end{aligned}$$
(a) Two conditions, two equations. A stationary point at $x=-1$ means $f'(-1)=0.$ The denominator is $(-3)^{2}=9\ne0,$ so the numerator must vanish: $$\begin{aligned} 1+4-2a-b&=0\,,\ \text{that is}\\ 2a+b&=5\,. \end{aligned}$$ And $f(-1)=-1$ means $\dfrac{1-a+b}{-3}=-1,$ that is $$\begin{aligned} 1-a+b&=3\,,\ \text{or}\\ -a+b&=2\,. \end{aligned}$$
(a) Solve. Subtracting the second equation from the first gives $3a=3,$ so $a=1,$ and then $b=2+a=3.$ Check: $f'(-1)$ has numerator $1+4-2-3=0,$ and $f(-1)=\dfrac{1-1+3}{-3}=-1.$
(b) The factored derivative. With $a=1$ and $b=3,$ $$\begin{aligned} f(x)&=\frac{x^{2}+x+3}{x-2}\,,\\ f'(x)&=\frac{x^{2}-4x-5}{(x-2)^{2}}\\ &=\frac{(x+1)(x-5)}{(x-2)^{2}}\,. \end{aligned}$$ $f'(x)=0$ at $x=-1$ and $x=5,$ and $f'$ is not defined at $x=2.$
(b) The sign table. $(x-2)^{2}>0$ for every $x\ne2,$ so the sign of $f'$ is the sign of $(x+1)(x-5).$ The pole $x=2$ gets a row of its own, where $f'$ is not defined (the dash).
x x+1 x−5 f′ (−∞,−1) − − + −1 0 − 0 (−1,2) + − − 2 + − — (2,5) + − − 5 + 0 0 (5,∞) + + + (b) The intervals. By Fact 2.23, on each interval inside the domain:
$f$ is increasing on $(-\infty,-1]$ and on $[5,\infty);$
$f$ is decreasing on $[-1,2)$ and on $(2,5].$
The two intervals of decrease are separate: $f$ is not defined at $2.$
(b) The largest value on $(-\infty,2).$ On the interval $I=(-\infty,2)$ we have $f'(x)\ge0$ for $x<-1$ and $f'(x)\le0$ for $-1<x<2.$ By the first-derivative test (Fact 6.18) the point $x=-1$ is a maximum point of $f$ over $I,$ and the largest value there is $f(-1)=-1.$
(b) The smallest value on $(2,\infty).$ On $(2,\infty)$ we have $f'(x)\le0$ for $2<x<5$ and $f'(x)\ge0$ for $x>5,$ so $x=5$ is a minimum point of $f$ over $(2,\infty),$ with $$f(5)=\frac{25+5+3}{3}=11\,.$$ (Lecture 7 calls $x=-1$ a local maximum point and $x=5$ a local minimum point of $f.$)
(c) Divide first. Division with remainder (Fact 1.22) gives $x^{2}+x+3=(x-2)(x+3)+9,$ so $$f(x)=x+3+\frac{9}{x-2}\,.$$ Check by multiplying back: $(x-2)(x+3)+9=x^{2}+x-6+9.$
(c) No maximum point. For $x>2$ the fraction $\dfrac{9}{x-2}$ is positive, so $f(x)>x+3.$ Given any number $M,$ every $x>\max(2,M)$ has $f(x)>M+3>M.$ So $f$ takes arbitrarily large values, and no value $f(c)$ can be the largest: $f$ has no maximum point.
(c) No minimum point. For $x<2$ the fraction is negative, so $f(x)<x+3,$ and $f$ takes arbitrarily large negative values in the same way: no minimum point either. Notice that the largest value on the left, $-1,$ is smaller than the smallest value on the right, $11.$ (That the domain is not a closed interval decides nothing: Watch out 6.10.)
Answer
(a) $a=1$ and $b=3.$
(b) $f'(x)=\dfrac{(x+1)(x-5)}{(x-2)^{2}}.$ Increasing on $(-\infty,-1]$ and on $[5,\infty),$ decreasing on $[-1,2)$ and on $(2,5].$ The largest value on $(-\infty,2)$ is $f(-1)=-1;$ the smallest value on $(2,\infty)$ is $f(5)=11.$
(c) No maximum point and no minimum point: $f(x)=x+3+\dfrac{9}{x-2}$ takes arbitrarily large values for $x>2$ and arbitrarily large negative values for $x<2.$
Mark your solution
Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
Your points 0 of 10 p
Common errors
- Setting the numerator equal to the value: $1-a+b=-1$ instead of $\dfrac{1-a+b}{-3}=-1.$ The condition is on $f(-1),$ and $f$ has a denominator.
- The numerator of the quotient rule in the wrong order, which reverses every sign in the table. In the practice problems on the quotient rule fewer than half of the answers were right; put $g\equiv1$ to check the order.
- “Decreasing on $[-1,5]$” — the interval runs across the pole. The function is not defined at $2,$ and $f(1)=-5<f(3)=15.$
- “The domain is not closed and bounded, so by the extreme value theorem $f$ has no maximum.” In the live vote, about a quarter of the answers accepted exactly this argument. It proves nothing (Watch out 6.10); the answer must come from the function.
A cubic with two parameters, then its extremes on a closed interval
For real numbers $a$ and $b,$ let $$f(x)=x^{3}+ax^{2}+bx+7\,.$$
- (a)3 p
Find $a$ and $b$ such that $f$ has a stationary point at $x=3$ with $f(3)=-20.$
- (b)2 p
With these values, determine the intervals on which $f$ is increasing and on which it is decreasing.
- (c)3 p
Find the largest and the smallest value of $f$ on the interval $[0,5],$ and where they are taken.
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(a) The derivative. $f'(x)=3x^{2}+2ax+b.$
(a) Two equations. $f'(3)=0$ gives $27+6a+b=0,$ that is $$6a+b=-27\,.$$ $f(3)=-20$ gives $27+9a+3b+7=-20,$ that is $9a+3b=-54,$ or $$3a+b=-18\,.$$
(a) Solve. Subtracting gives $3a=-9,$ so $a=-3,$ and $b=-18-3a=-9.$ Check: $f'(3)=27-18-9=0$ and $f(3)=27-27-27+7=-20.$
(b) The sign of $f'.$ With $a=-3,$ $b=-9{:}$ $$\begin{aligned} f'(x)&=3x^{2}-6x-9\\ &=3(x+1)(x-3)\,. \end{aligned}$$
x x+1 x−3 f′ (−∞,−1) − − + −1 0 − 0 (−1,3) + − − 3 + 0 0 (3,∞) + + + (b) The intervals. By Method 2.25, $f$ is increasing on $(-\infty,-1]$ and on $[3,\infty),$ and decreasing on $[-1,3].$
(c) The guarantee. $f$ is a polynomial, hence continuous, and $[0,5]$ is closed and bounded, so by Fact 6.9 the largest and the smallest value exist.
(c) The candidates. $f$ is differentiable everywhere, so the candidates are the stationary points inside $[0,5]$ and the endpoints. Of $-1$ and $3,$ only $3$ lies in $[0,5].$ The candidates are $0,$ $3$ and $5{:}$ $$\begin{aligned} f(0)&=7\,,\\ f(3)&=-20\,,\\ f(5)&=125-75-45+7\\ &=12\,. \end{aligned}$$
(c) Compare. The largest value is $12,$ at the endpoint $x=5,$ where $f'(5)=36\ne0.$ The smallest value is $-20,$ at the stationary point $x=3.$
Answer
(a) $a=-3$ and $b=-9.$
(b) Increasing on $(-\infty,-1]$ and on $[3,\infty);$ decreasing on $[-1,3].$
(c) On $[0,5]$ the largest value is $f(5)=12$ and the smallest is $f(3)=-20.$
Mark your solution
Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
Your points 0 of 8 p
Common errors
- Taking the stationary points as the answer: “the largest value is $f(-1)=12$ at $x=-1$”. The value happens to be right, but $x=-1$ lies outside $[0,5],$ and the maximum is at the endpoint $x=5.$ In the practice problems, the template in which the stationary point may lie outside the interval was answered correctly only $42\%$ of the time.
- Not evaluating the endpoints at all. In the live vote on the same question for a parabola, about one answer in six put the maximum at the stationary point (Watch out 6.16).
- Using only $f'(3)=0,$ which is one equation for two unknowns. The value $f(3)=-20$ is the second equation.
- Forgetting the constant $7$ in $f(3)=-20,$ which gives $9a+3b=-47$ and fractions.
Exactly one solution: the argument behind “the only possible value”
(a) Show that the equation $$e^{y}+y=2$$ has exactly one real solution, and that it lies between $0$ and $1.$
(b) The equation $$x\,e^{y}+3y=2x-1$$ defines $y$ as a differentiable function of $x$ near $x=1.$ Find all possible values of $y(1).$
- (a)3 p
Exactly one solution of $e^{y}+y=2,$ in $(0,1).$
- (b)2 p
All possible values of $y(1).$
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(a) At least one. Put $h(y)=e^{y}+y-2.$ It is continuous, and $$\begin{aligned} h(0)&=1+0-2=-1<0\,,\\ h(1)&=e+1-2=e-1>0\,. \end{aligned}$$ By the intermediate value theorem (Fact 6.1) $h$ has a zero in $(0,1).$
(a) At most one. $h'(y)=e^{y}+1>0$ for every $y,$ so $h$ is strictly increasing (Fact 2.23) and takes the value $0$ at most once.
(a) Exactly one. Together: the equation has exactly one real solution, and it lies in $(0,1).$ (Two separate tasks, as in Example 6.4; the sign change alone gives only “at least one”, Watch out 6.2.)
(b) Insert $x=1.$ The equation becomes $e^{y}+3y=2-1,$ that is $$e^{y}+3y=1\,.$$ $y=0$ is a solution: $e^{0}+0=1.$
(b) The only one. $g(y)=e^{y}+3y$ has $g'(y)=e^{y}+3>0,$ so $g$ is strictly increasing and takes the value $1$ only once. Hence $y(1)=0$ is the only possible value.
Answer
(a) $h(y)=e^{y}+y-2$ has $h(0)<0<h(1)$ and $h'(y)=e^{y}+1>0{:}$ exactly one solution, in $(0,1).$
(b) $y(1)=0$ is the only possible value.
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Common errors
- Stopping after the sign change: “$h(0)<0<h(1),$ so there is exactly one solution”. The intermediate value theorem gives at least one (Watch out 6.2); exactly one needs the second argument.
- Answering (b) with “$y=0$” and no word about other values. The problem asks for all possible values, and the exam solutions require the reason why there are no others.
- Citing the intermediate value theorem without saying that $h$ is continuous — the one hypothesis it has. Without continuity a sign change proves nothing, as $1/x$ on $[-1,1]$ shows (given any value at $0$).