7 Problems
Shape, and then turning round
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.
Typing an answer. Write powers with ^ or ** and logarithms as ln(x+2): for example x^2/2 - 3x + 4 ln(x+2) - 4 ln(2). The preview under the box shows how your answer was read. Look at it before you press Check. An antiderivative asked for here always has its constant fixed by a condition, so there is no $+C$ to type.
●○○ Basic●●○ Intermediate●●● Challenge
Checked instantlySolution included
Local extreme points
Read the theory: 1. Local extreme points →Let $f(x)=x^{4}-8x^{2}+3.$
(a) Find every $x$ with $f'(x)=0.$
(b) Classify each of them with the second-derivative test, and give the value of $f$ there.
(c) Is the local maximum point a maximum point of $f$ over the whole real line? Compute $f(3)$ before you answer.
Hint
Take out the common factor $4x$ from $f'(x).$ For (c), remember Watch out 7.2: a local maximum only beats its neighbours.
(a) The candidates. $$ \begin{aligned} f'(x)&=4x^{3}-16x\\ &=4x(x^{2}-4)\\ &=4x(x-2)(x+2)\,, \end{aligned} $$ so $f'(x)=0$ at $x=-2,$ $x=0$ and $x=2.$
(b) The second-derivative test. $f''(x)=12x^{2}-16.$ $$ \begin{aligned} f''(-2)&=48-16=32>0\,,\\ f''(0)&=-16<0\,,\\ f''(2)&=32>0\,. \end{aligned} $$ By Fact 7.3: local minima at $x=\pm2$ and a local maximum at $x=0.$ The values are $f(0)=3$ and $f(\pm2)=16-32+3=-13.$
(c) Local, not global. $f(3)=81-72+3=12,$ which is larger than $f(0)=3.$ So $x=0$ is not a maximum point over the whole line: it beats its neighbours and nothing more.
Answer
(a) $x=-2,$ $0,$ $2.$
(b) Local maximum at $x=0,$ $f(0)=3;$ local minima at $x=\pm2,$ $f(\pm2)=-13.$
(c) No: $f(3)=12>3.$
How did it go?
Stuck? Read 1. Local extreme points →
Inflection points
Read the theory: 2. Inflection points →A workshop with $L$ workers ($0\le L\le6$) produces $$Q(L)=6L^{2}-L^{3}$$ units a day.
(a) Find the marginal product $Q'(L).$
(b) On which interval does the marginal product rise, and on which does it fall? Where does $Q$ have an inflection point?
(c) For which $L$ is the output largest? Compare with the endpoints.
(d) Say in words what happens to the output between $L=2$ and $L=4,$ and after $L=4.$
Hint
The marginal product rises where its own derivative $Q''(L)$ is positive. An inflection point needs a change of sign of $Q''$ (Method 7.10).
(a) $Q'(L)=12L-3L^{2}=3L(4-L).$
(b) The sign of $Q''.$ $Q''(L)=12-6L=6(2-L).$
L<2 2 L>2 ${Q''(L)}$ + 0 − The marginal product rises on $[0,2]$ and falls on $[2,6].$ $Q''$ changes sign at $L=2,$ so $L=2$ is an inflection point of $Q,$ with $Q(2)=24-8=16.$ There the marginal product is largest: $Q'(2)=12.$
(c) The largest output. $Q'(L)=0$ for $L=0$ and $L=4.$ On $[0,6]$ the candidates are $L=0,$ $4,$ $6$ (Method 6.15): $$ \begin{aligned} Q(0)&=0\,,\\ Q(4)&=96-64=32\,,\\ Q(6)&=216-216=0\,. \end{aligned} $$ The output is largest, $32$ units, with $L=4$ workers.
(d) In words. Up to $L=2$ each extra worker adds more than the one before. Between $L=2$ and $L=4$ the output still rises, but each extra worker adds less (diminishing returns: the inflection point is where they set in). After $L=4$ an extra worker lowers the output: $Q'(L)<0.$
Answer
(a) $Q'(L)=12L-3L^{2}.$
(b) Rising on $[0,2],$ falling on $[2,6];$ inflection point at $L=2,$ where $Q=16.$
(c) At $L=4,$ with $Q(4)=32.$
(d) Diminishing returns from $L=2;$ falling output after $L=4.$
How did it go?
Stuck? Read 2. Inflection points →
Turning round: antiderivatives
Read the theory: 3. Turning round: antiderivatives →A firm's marginal cost is $$C'(Q)=3Q^{2}-12Q+20\,,$$ and producing $2$ units costs $100$ in total.
(a) Find the total cost function $C(Q).$
(b) What is the fixed cost?
(c) For which $Q$ is the marginal cost smallest, and what is it there? What does the cost curve $C(Q)$ do at that $Q$?
Hint
Integrate term by term and keep the constant; then use $C(2)=100.$ In (c), the marginal cost is a function of its own: find where its derivative $C''(Q)$ is zero and changes sign.
(a) Integrate, and keep the constant. $$ \begin{aligned} &C(Q)\\ &=\int(3Q^{2}-12Q+20)\,dQ\\ &=Q^{3}-6Q^{2}+20Q+K\,. \end{aligned} $$ Check: differentiating gives $3Q^{2}-12Q+20$ back.
The constant from the condition. $C(2)=8-24+40+K=24+K,$ and this is $100,$ so $K=76.$ Hence $$C(Q)=Q^{3}-6Q^{2}+20Q+76\,.$$
(b) The fixed cost is $C(0)=K=76$ (Example 7.15).
(c) The smallest marginal cost. $C''(Q)=6Q-12=6(Q-2),$ negative for $Q<2$ and positive for $Q>2.$ So $C'$ falls until $Q=2$ and rises after it: the marginal cost is smallest at $Q=2,$ where $C'(2)=12-24+20=8.$
What the cost curve does there. $C''$ changes sign at $Q=2,$ so $Q=2$ is an inflection point of $C{:}$ the cost curve bends downwards before it and upwards after it. Costs rise ever more slowly up to $Q=2$ and ever faster beyond.
Answer
(a) $C(Q)=Q^{3}-6Q^{2}+20Q+76.$
(b) $76.$
(c) At $Q=2,$ where $C'(2)=8;$ $Q=2$ is an inflection point of the cost curve.
How did it go?
Stuck? Read 3. Turning round: antiderivatives →
(a) Show that $F(x)=\ln(2x)$ and $G(x)=\ln x$ are both antiderivatives of $\dfrac1x$ for $x>0.$ What is $F-G,$ and why does that agree with Fact 7.13?
(b) For $x\ne0$ let $$ H(x)=\begin{cases}\ln|x|+1, & x>0\,,\\ \ln|x|-2, & x<0\,.\end{cases} $$ Show that $H'(x)=\dfrac1x$ for every $x\ne0.$ Is $H(x)-\ln|x|$ constant? Explain why this does not contradict Fact 7.13.
Hint
In (a), use the chain rule, or first write $\ln(2x)=\ln2+\ln x.$ In (b), for $x<0$ we have $|x|=-x.$ Then read the hypothesis of Fact 7.13 again, and Watch out 7.14.
(a) Both are antiderivatives. By the chain rule $F'(x)=\dfrac{1}{2x}\cdot2=\dfrac1x,$ and $G'(x)=\dfrac1x.$ Their difference is $$ \begin{aligned} F(x)-G(x)&=\ln(2x)-\ln x\\ &=\ln2\,, \end{aligned} $$ a constant, as Fact 7.13 says it must be on the interval $x>0.$
(b) The derivative of $H.$ For $x>0,$ $H(x)=\ln x+1$ and $H'(x)=\dfrac1x.$ For $x<0,$ $H(x)=\ln(-x)-2,$ and by the chain rule $H'(x)=\dfrac{1}{-x}\cdot(-1)=\dfrac1x.$
The difference is not constant. $H(x)-\ln|x|$ is $1$ for $x>0$ and $-2$ for $x<0.$ It has two values.
Why there is no contradiction. Fact 7.13 assumes $F'=G'$ on an interval. The set $x\ne0$ is not an interval: it is two intervals with a gap at $0.$ On each of them separately the difference is constant ($1$ on the right, $-2$ on the left), exactly as the Fact says. The constant may jump across the gap, because nothing connects the two pieces.
What this says about $\int\frac1x\,dx=\ln|x|+C.$ On $x\ne0$ the constant may be chosen separately on $x>0$ and on $x<0.$ In practice a problem states the interval (as the exam does: “for $x>1$”), and then one $C$ is all the freedom there is.
Answer
(a) $F'=G'=\dfrac1x$ and $F-G=\ln2,$ a constant on the interval $x>0.$
(b) $H'(x)=\dfrac1x$ for all $x\ne0,$ but $H-\ln|x|$ is $1$ for $x>0$ and $-2$ for $x<0.$ No contradiction: $x\ne0$ is not an interval, and on each of its two intervals the difference is constant.
How did it go?
Stuck? Read 3. Turning round: antiderivatives →
The surprise: it measures area
Read the theory: 4. The surprise: it measures area →Let $f(x)=x^{2}-4.$
(a) Compute $\displaystyle\int_{0}^{3}f(x)\,dx.$
(b) Compute the area of the region between the graph of $f$ and the $x$-axis for $0\le x\le3.$
(c) Why are the two answers different, and why is the first one negative?
Hint
Find where $f$ changes sign on $[0,3].$ The area is the sum of the areas on each side of that point, each counted as positive (Watch out 7.19).
(a) The integral. An antiderivative is $F(x)=\tfrac13x^{3}-4x,$ so $$ \begin{aligned} &\int_{0}^{3}(x^{2}-4)\,dx\\ &=F(3)-F(0)\\ &=(9-12)-0=-3\,. \end{aligned} $$
(b) Split at the root. $f(x)=(x-2)(x+2)$ is negative on $[0,2)$ and positive on $(2,3].$ By Fact 7.18, $$ \begin{aligned} \int_{0}^{2}f&=F(2)-F(0)\\ &=\tfrac83-8=-\tfrac{16}{3}\,, \end{aligned} $$ $$ \begin{aligned} \int_{2}^{3}f&=F(3)-F(2)\\ &=-3+\tfrac{16}{3}=\tfrac73\,. \end{aligned} $$ The area counts both parts as positive: $$\tfrac{16}{3}+\tfrac73=\tfrac{23}{3}\,.$$
(c) Signed area. The integral counts the part below the axis as negative: $-\tfrac{16}{3}+\tfrac73=-3.$ The part below the axis ($\tfrac{16}{3}$) is larger than the part above it ($\tfrac73$), so the integral is negative. Only when $f\ge0$ throughout is the integral the area.
Answer
(a) $-3.$
(b) $\tfrac{23}{3}.$
(c) The integral is signed area: $-\tfrac{16}{3}+\tfrac73=-3,$ while the area is $\tfrac{16}{3}+\tfrac73.$
How did it go?
Stuck? Read 4. The surprise: it measures area →
For $0\le t\le8$ let $$A(t)=\int_{0}^{t}(4-x)\,dx\,.$$
(a) Compute $A(t)$ with Definition 7.16, and find $A'(t).$ Compare with the function under the integral sign.
(b) On which interval is $A$ increasing, and on which decreasing? Where is $A$ largest, and what is its largest value?
(c) Find the $t>0$ with $A(t)=0,$ and explain the answer with a picture of the region.
Hint
An antiderivative of $4-x$ is $4x-\tfrac12x^{2}.$ In (b), the sign of $A'$ decides (Method 7.10). In (c), think of signed area.
(a) The area function. With $F(x)=4x-\tfrac12x^{2},$ $$ \begin{aligned} A(t)&=F(t)-F(0)\\ &=4t-\tfrac12t^{2}\,, \end{aligned} $$ so $A'(t)=4-t.$ That is the function under the integral sign, evaluated at $t,$ as Fact 7.17 says.
(b) Where $A$ rises. $A'(t)=4-t$ is positive for $t<4$ and negative for $t>4.$ So $A$ increases on $[0,4]$ and decreases on $[4,8],$ and by the first-derivative test of Lecture 6 its largest value on $[0,8]$ is $$A(4)=16-8=8\,.$$ In words: up to $t=4$ the area arrives with positive height; after that $f(x)=4-x$ is negative and every new strip counts as negative.
(c) Back to zero. $A(t)=t\bigl(4-\tfrac12t\bigr)=0$ for $t=0$ and $t=8.$ The region from $0$ to $4$ is a triangle above the axis with area $\tfrac12\cdot4\cdot4=8;$ the region from $4$ to $8$ is the same triangle below the axis, which counts as $-8.$ They cancel at $t=8.$
Answer
(a) $A(t)=4t-\tfrac12t^{2}$ and $A'(t)=4-t.$
(b) Increasing on $[0,4],$ decreasing on $[4,8];$ the largest value is $A(4)=8.$
(c) $t=8{:}$ the triangle below the axis cancels the one above.
How did it go?
Stuck? Read 4. The surprise: it measures area →
Integration by parts
Read the theory: 5. Integration by parts →(a) Find $\displaystyle\int x^{2}e^{x}\,dx.$ You will need integration by parts twice.
(b) Check your answer by differentiating it.
(c) Why is $u=x^{2}$ the right choice, and not $u=e^{x}$?
Hint
Take $u=x^{2},$ $v'=e^{x}.$ The new integral is $\int2x\,e^{x}\,dx,$ which is Example 7.23 times $2.$
(a) Once. With $u=x^{2},$ $v'=e^{x},$ so $u'=2x$ and $v=e^{x},$ formula (7.2) gives $$ \begin{aligned} \int x^{2}e^{x}\,dx&=x^{2}e^{x}\\ &\quad-\int2x\,e^{x}\,dx\,. \end{aligned} $$
Twice. The new integral is $2$ times Example 7.23: $\int xe^{x}\,dx=xe^{x}-e^{x}+C.$ So $$ \begin{aligned} &\int x^{2}e^{x}\,dx\\ &=x^{2}e^{x}-2\bigl(xe^{x}-e^{x}\bigr)+C\\ &=(x^{2}-2x+2)e^{x}+C\,. \end{aligned} $$ The minus sign stands in front of the whole bracket, so $-e^{x}$ becomes $+2e^{x}.$
(b) The check. By the product rule, $$ \begin{aligned} &\bigl((x^{2}-2x+2)e^{x}\bigr)'\\ &=(2x-2)e^{x}\\ &\quad+(x^{2}-2x+2)e^{x}\\ &=x^{2}e^{x}\,. \end{aligned} $$
(c) The choice. Each differentiation lowers the power of $x$ by one, and $e^{x}$ stays $e^{x}$ when it is integrated: after two rounds the polynomial is a constant. With $u=e^{x}$ and $v'=x^{2}$ the new integral would be $\int e^{x}\tfrac{x^{3}}{3}\,dx,$ worse than the one we started with (Method 7.22).
Answer
(a) $(x^{2}-2x+2)e^{x}+C.$
(b) The derivative is $x^{2}e^{x}.$
(c) Differentiating $x^{2}$ twice leaves a constant; with $u=e^{x}$ the power of $x$ would grow.
How did it go?
Stuck? Read 5. Integration by parts →
Mixed, as on the exam
Here the sections are mixed, and nothing says which method to use. Write the solution on paper, with reasons, before you look.
Let $f(x)=x^{4}-4x^{3}.$
(a) Find the points where $f'(x)=0$ and classify each one. Where the second-derivative test gives no information, say so and use another way.
(b) Find the inflection points of $f.$
(c) One point appears in both answers. What does the curve look like there?
Hint
$f'(x)=4x^{2}(x-3)$ and $f''(x)=12x(x-2).$ The factor $4x^{2}$ is never negative (Watch out 7.4, Method 7.10).
(a) The candidates. $f'(x)=4x^{3}-12x^{2}=4x^{2}(x-3),$ zero at $x=0$ and $x=3.$ With $f''(x)=12x^{2}-24x,$ $$ \begin{aligned} f''(3)&=108-72=36>0\,,\\ f''(0)&=0\,. \end{aligned} $$ So $x=3$ is a local minimum, with $f(3)=81-108=-27.$ At $x=0$ the test is silent.
(a) The sign of $f'$ at $0.$ The factor $4x^{2}$ is positive on both sides of $0,$ so the sign of $f'$ is the sign of $x-3{:}$
${f'(x)}$ x < 0 − 0 < x < 3 − x > 3 + No change at $0{:}$ $x=0$ is neither a maximum nor a minimum. From $-$ to $+$ at $3,$ as the second-derivative test said.
(b) Inflection points. $f''(x)=12x(x-2)$ is zero at $0$ and $2.$
${f''(x)}$ x < 0 + 0 < x < 2 − x > 2 + The sign changes at both, so $x=0$ and $x=2$ are inflection points, with $f(0)=0$ and $f(2)=16-32=-16.$
(c) A flat inflection. At $x=0$ the tangent is horizontal ($f'(0)=0$) and the bending changes: the curve comes down, levels off for a moment and goes on down, like $-x^{3}$ at $0.$ At $x=2$ the inflection has a sloping tangent, $f'(2)=32-48=-16.$
Answer
(a) $x=3$ is a local minimum, $f(3)=-27;$ $x=0$ is neither ($f''(0)=0,$ and $f'$ is negative on both sides).
(b) $x=0$ and $x=2.$
(c) $x=0{:}$ a horizontal tangent at an inflection point.
How did it go?
Stuck? Read 1. Local extreme points →
Compute $$\int_{0}^{2}\frac{x^{2}+1}{x+1}\,dx\,.$$ Give the exact value and a decimal approximation, and check that the size is reasonable.
Hint
Divide first (Fact 1.22), then integrate the polynomial and the remainder over $x+1,$ and use Definition 7.16.
Divide. $(x+1)(x-1)=x^{2}-1,$ so the remainder is $2{:}$ $$\frac{x^{2}+1}{x+1}=x-1+\frac{2}{x+1}\,.$$ Check: $(x+1)(x-1)+2=x^{2}+1.$
An antiderivative. On $[0,2]$ we have $x+1>0,$ so $$F(x)=\tfrac12x^{2}-x+2\ln(x+1)\,.$$ Its derivative is $x-1+\dfrac{2}{x+1}.$
Evaluate. $$ \begin{aligned} &F(2)-F(0)\\ &=(2-2+2\ln3)-0\\ &=2\ln3\approx2.20\,. \end{aligned} $$
Sanity check. On $[0,2]$ the integrand is positive, and it is at most $x+1,$ because $x^{2}+1\le(x+1)^{2}$ for $x\ge0.$ So the integral lies between $0$ and $\int_{0}^{2}(x+1)\,dx=4.$ The value $2.20$ does.
Answer
$2\ln3\approx2.20.$
How did it go?
Stuck? Read 3. Turning round: antiderivatives →
As on the exam
4 problems · 26 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.
Divide first, then integrate
Calculate $$\int \frac{x^{3}+x^{2}-4x+10}{x+3}\,dx$$ for $x>-3.$
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Why we divide. The numerator has degree $3$ and the denominator degree $1,$ so the integrand is an improper rational function. By Fact 1.22 it is a polynomial plus a remainder over $x+3,$ and each of those has an antiderivative we know.
The division. Long division (or matching coefficients) gives the quotient $x^{2}-2x+2$ and the remainder $4{:}$ $$ \begin{aligned} &\frac{x^{3}+x^{2}-4x+10}{x+3}\\ &\quad=x^{2}-2x+2+\frac{4}{x+3}\,. \end{aligned} $$
Check by multiplying back. $$ \begin{aligned} &(x+3)(x^{2}-2x+2)\\ &=x^{3}-2x^{2}+2x\\ &\quad+3x^{2}-6x+6\\ &=x^{3}+x^{2}-4x+6\,, \end{aligned} $$ and adding the remainder $4$ gives $x^{3}+x^{2}-4x+10.$
The polynomial. By the power rule backwards (Fact 7.13), $$ \begin{aligned} &\int(x^{2}-2x+2)\,dx\\ &=\tfrac13x^{3}-x^{2}+2x+C\,. \end{aligned} $$
The remainder. The derivative of $\ln|x+3|$ is $\dfrac{1}{x+3},$ and for $x>-3$ we have $x+3>0,$ so $\ln|x+3|=\ln(x+3)$ there. Hence $$\int\frac{4}{x+3}\,dx=4\ln(x+3)+C\,.$$
Together, for $x>-3,$ $$ \begin{aligned} &\int \frac{x^{3}+x^{2}-4x+10}{x+3}\,dx\\ &=\tfrac13x^{3}-x^{2}+2x\\ &\quad+4\ln(x+3)+C\,, \end{aligned} $$ where $C$ is an arbitrary constant. Differentiating gives $x^{2}-2x+2+\dfrac{4}{x+3}$ back.
Answer
$\tfrac13x^{3}-x^{2}+2x+4\ln(x+3)+C$ for $x>-3,$ with $C$ an arbitrary constant.
Mark your solution
Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
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Common errors
- The constant term of the numerator ($10$) taken as the remainder, which gives $10\ln(x+3).$ The remainder is what is left after the division: $4.$ The check by multiplying back catches it.
- The numerator and the denominator integrated separately, $\bigl(\tfrac14x^{4}+\dots\bigr)\big/\bigl(\tfrac12x^{2}+3x\bigr).$ There is no quotient rule for integrals; divide first.
- A sign slip in the long division ($x^{2}-3x^{2}$ taken as $+2x^{2}$ or $4x^{2}$ instead of $-2x^{2}$), so that the quotient is wrong. Multiplying back finds it at once.
- The $+C$ left out. The exam asks for the indefinite integral, which is a whole family of functions.
By parts: a logarithm, and the base 3
Calculate the integrals
- (a)3 p$\displaystyle\int_{1}^{e} x^{3}\ln x\,dx,$
- (b)2 p$\displaystyle\int x\cdot3^{x}\,dx.$
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(a) Which factor is $u.$ We have no antiderivative of $\ln x,$ but its derivative is $\dfrac1x$ (Fact 4.5); $x^{3}$ is easy to integrate. So, by Method 7.22 as in Example 7.24, $$ \begin{aligned} u&=\ln x\,,& v'&=x^{3}\,,\\ u'&=\frac1x\,,& v&=\frac{x^{4}}{4}\,. \end{aligned} $$
(a) An antiderivative. By (7.2), $$ \begin{aligned} &\int x^{3}\ln x\,dx\\ &=\frac{x^{4}}{4}\ln x-\int\frac{x^{4}}{4}\cdot\frac1x\,dx\\ &=\frac{x^{4}}{4}\ln x-\int\frac{x^{3}}{4}\,dx\\ &=\frac{x^{4}}{4}\ln x-\frac{x^{4}}{16}+C\,. \end{aligned} $$ Check: the derivative of $F(x)=\frac{x^{4}}{4}\ln x-\frac{x^{4}}{16}$ is $x^{3}\ln x+\frac{x^{3}}{4}-\frac{x^{3}}{4}=x^{3}\ln x.$
(a) Evaluate. With $\ln e=1$ and $\ln1=0,$ $$ \begin{aligned} F(e)&=\frac{e^{4}}{4}-\frac{e^{4}}{16}=\frac{3e^{4}}{16}\,,\\ F(1)&=0-\frac{1}{16}=-\frac{1}{16}\,, \end{aligned} $$ so by Definition 7.16 $$ \begin{aligned} &\int_{1}^{e}x^{3}\ln x\,dx\\ &=\frac{3e^{4}}{16}+\frac{1}{16}\\ &=\frac{3e^{4}+1}{16}\approx10.30\,. \end{aligned} $$ Sanity check: the integrand is $\ge0$ on $[1,e]$ and at most $e^{3}\approx20.1,$ and the interval has length $e-1\approx1.72,$ so the value must lie between $0$ and about $34.5.$
(b) An antiderivative of $3^{x}.$ Since $3^{x}=e^{x\ln3},$ its derivative is $3^{x}\ln3$ (Fact 4.4). Dividing by the constant $\ln3,$ $$\int3^{x}\,dx=\frac{3^{x}}{\ln3}+C\,.$$
(b) By parts. The polynomial $x$ simplifies when differentiated, so $u=x,$ $v'=3^{x},$ $u'=1,$ $v=\dfrac{3^{x}}{\ln3}{:}$ $$ \begin{aligned} &\int x\,3^{x}\,dx\\ &=\frac{x\,3^{x}}{\ln3}-\int\frac{3^{x}}{\ln3}\,dx\\ &=\frac{x\,3^{x}}{\ln3}-\frac{3^{x}}{(\ln3)^{2}}+C\,. \end{aligned} $$
(b) Check. By the product rule the derivative is $$ \begin{aligned} &\frac{3^{x}}{\ln3}+\frac{x\,3^{x}\ln3}{\ln3}-\frac{3^{x}\ln3}{(\ln3)^{2}}\\ &=x\,3^{x}\,. \end{aligned} $$
Answer
(a) $\dfrac{3e^{4}+1}{16}\approx10.30.$
(b) $\dfrac{x\,3^{x}}{\ln3}-\dfrac{3^{x}}{(\ln3)^{2}}+C.$
Mark your solution
Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
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Common errors
- In (a), $u=x^{3}$ and $v'=\ln x{:}$ the rule of thumb “the polynomial is $u$” applied against a logarithm. It is about polynomials against exponentials; we cannot integrate $\ln x,$ so it must be $u$ (Example 7.24).
- The value at the lower limit: $F(1)=-\tfrac{1}{16},$ and subtracting it adds $\tfrac{1}{16}.$ Writing $\tfrac{3e^{4}-1}{16}$ loses that sign.
- In (b), $\int3^{x}\,dx=\frac{3^{x+1}}{x+1}{:}$ the power rule used where the variable is in the exponent. Write $3^{x}=e^{x\ln3}$ first.
- In (b), only one factor of $\ln3$ in the last term: $\int\frac{3^{x}}{\ln3}\,dx$ brings a second one, giving $(\ln3)^{2}.$
A polynomial times $e^{-x}$: parameters, extrema, and global extremes
For real numbers $a$ and $b,$ consider the function $$f(x)=\bigl(x^{2}+ax+b\bigr)e^{-x}\,.$$
- (a)3 p
Find the values of $a$ and $b$ such that $f$ has stationary points at both $x=-1$ and $x=2.$
- (b)4 p
For the values of $a$ and $b$ found in (a), determine the intervals on which $f$ is increasing and on which it is decreasing, and classify the stationary points as local maximum or minimum points. Give the value of $f$ at each.
- (c)3 p
Does $f$ have a global maximum on $\mathbb{R}$? A global minimum? Justify your answers.
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(a) The derivative. By the product rule, with $(e^{-x})'=-e^{-x},$ $$ \begin{aligned} f'(x)&=(2x+a)e^{-x}\\ &\quad-(x^{2}+ax+b)e^{-x}\\ &=\bigl(-x^{2}+(2-a)x\\ &\qquad+a-b\bigr)e^{-x}\,. \end{aligned} $$
(a) The two conditions. Since $e^{-x}>0,$ $f'(x)=0$ exactly when the bracket is $0.$ At $x=-1$ and at $x=2{:}$ $$ \begin{aligned} -1-(2-a)+a-b&=0\,,\\ -4+2(2-a)+a-b&=0\,, \end{aligned} $$ that is $2a-b=3$ and $a+b=0.$
(a) Solve. Adding the two equations gives $3a=3,$ so $a=1$ and $b=-1.$ Check: the bracket becomes $-x^{2}+x+2=-(x+1)(x-2),$ which is zero at $-1$ and $2.$
(b) The sign of $f'.$ Now $f(x)=(x^{2}+x-1)e^{-x}$ and $$f'(x)=-(x+1)(x-2)\,e^{-x}\,.$$ The factor $e^{-x}$ is positive, so the sign of $f'$ is the sign of $-(x+1)(x-2){:}$
${x+1}$ ${x-2}$ ${f'}$ x < −1 − − − −1 < x < 2 + − + x > 2 + + − $f'(-1)=f'(2)=0.$(b) The intervals. By Fact 2.23, $f$ is decreasing on $(-\infty,-1]$ and on $[2,\infty),$ and increasing on $[-1,2].$
(b) Classify. $f'$ goes from $-$ to $+$ at $x=-1{:}$ a local minimum point, with $$f(-1)=(1-1-1)e^{1}=-e\,.$$ $f'$ goes from $+$ to $-$ at $x=2{:}$ a local maximum point, with $$ \begin{aligned} f(2)&=(4+2-1)e^{-2}\\ &=5e^{-2}\approx0.68\,. \end{aligned} $$
(c) No global maximum. For a natural number $n\ge2,$ $$ \begin{aligned} f(-n)&=(n^{2}-n-1)e^{n}\\ &\ge e^{n}\,, \end{aligned} $$ since $n^{2}-n-1\ge1$ for $n\ge2.$ So $f$ takes values larger than any number we name, and there is no largest value. (The local maximum value $5e^{-2}$ is only the largest value on $[-1,\infty).$)
(c) A global minimum at $x=-1.$ On $(-\infty,-1]$ $f$ is decreasing, so $f(x)\ge f(-1)$ there. On $[-1,2]$ it is increasing, so $f(x)\ge f(-1)$ there too. For $x\ge2$ we have $x^{2}+x-1\ge5>0,$ so $f(x)>0>-e.$ Hence $f(x)\ge f(-1)=-e$ for every $x{:}$ $x=-1$ is a global minimum point, and the minimum value is $-e.$
Answer
(a) $a=1,$ $b=-1.$
(b) Decreasing on $(-\infty,-1]$ and on $[2,\infty),$ increasing on $[-1,2].$ Local minimum at $x=-1$ with $f(-1)=-e;$ local maximum at $x=2$ with $f(2)=5e^{-2}.$
(c) No global maximum. Global minimum $-e,$ at $x=-1.$
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Common errors
- $(e^{-x})'=e^{-x}{:}$ the inner derivative $-1$ of the chain rule forgotten. Then $f'=(x^{2}+(a+2)x+a+b)e^{-x}$ and the system gives other values of $a$ and $b.$
- The minus sign in front of the bracket lost, so that the sign table is upside down and the local maximum and minimum are swapped. Check with two values: $f(-1)=-e$ is negative and $f(2)=5e^{-2}$ is positive, so $-1$ cannot be the maximum.
- “$f(2)=5e^{-2}$ is the global maximum.” A local maximum is compared with its neighbours only; for $x<-1$ the values grow without bound.
- The global minimum concluded from “$-e$ is the smaller of the two stationary values” without looking at the rest of the line. The argument must cover every $x.$
Where a rational function and a cubic are convex
Determine the intervals on which the function is convex and those on which it is concave, and find all inflection points.
- (a)3 p
$$f(x)=\frac{x^{2}+3x+1}{x-1}$$ for $x\ne1.$
- (b)3 p$$g(x)=-x^{3}+3x^{2}+9x-4\,.$$
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(a) Divide first. The numerator has the higher degree, and dividing makes $f''$ easy. $(x-1)(x+4)=x^{2}+3x-4,$ so the remainder is $5{:}$ $$f(x)=x+4+\frac{5}{x-1}\,.$$ Check: $(x-1)(x+4)+5=x^{2}+3x+1.$
(a) Two derivatives. With $\dfrac{5}{x-1}=5(x-1)^{-1},$ $$ \begin{aligned} f'(x)&=1-\frac{5}{(x-1)^{2}}\,,\\ f''(x)&=\frac{10}{(x-1)^{3}}\,. \end{aligned} $$
(a) The sign of $f''.$ $(x-1)^{3}$ has the sign of $x-1{:}$
x<1 x>1 ${f''(x)}$ − + ($f''$ is not defined at $x=1.$)
By Fact 6.20, $f$ is concave on $(-\infty,1)$ and convex on $(1,\infty).$ There is no inflection point: the bending changes across $x=1,$ but $1$ is not in the domain (Definition 7.6 needs a point of the domain).
(b) The second derivative. $g'(x)=-3x^{2}+6x+9$ and $$g''(x)=-6x+6=-6(x-1)\,.$$
(b) The sign of $g''.$ Because of the factor $-6,$ $g''$ is positive for $x<1$ and negative for $x>1{:}$
x<1 1 x>1 ${g''(x)}$ + 0 − So $g$ is convex on $(-\infty,1]$ and concave on $[1,\infty).$
(b) The inflection point. $g''(1)=0$ and $g''$ changes sign at $x=1$ (Method 7.10), so $x=1$ is an inflection point. It is the point $(1,g(1))=(1,7),$ since $g(1)=-1+3+9-4=7.$
Answer
(a) Concave on $(-\infty,1),$ convex on $(1,\infty);$ no inflection point.
(b) Convex on $(-\infty,1],$ concave on $[1,\infty);$ inflection point $(1,7).$
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Common errors
- In (a), $x=1$ called an inflection point because the bending changes there. $f$ is not defined at $1,$ and an inflection point is a point of the domain.
- In (a), the quotient rule used twice on $\frac{x^{2}+3x+1}{x-1}$ and the numerator of $f''$ not simplified, so that its sign cannot be read. Dividing first leaves $\frac{10}{(x-1)^{3}}.$
- In (b), convex and concave swapped: $-6x+6$ is positive to the left of $1.$
- An inflection point from $g''(1)=0$ alone (Watch out 7.8). Here the sign does change, but the solution has to say so.