Revision: derivatives and graphs of functions | Grade 12

This topic usually appears in all three parts of the paper: a few short questions in Part I (reading a table of variations, finding asymptotes), usually a true/false question about one function in Part II, and one or two maximum/minimum questions in Part III, altogether usually worth about $1.5$ to $2.5$ marks.

For the details, look back at the lessons on turning points, maximum and minimum values, asymptotes and curve sketching.

Monotonicity and turning points

Typical questions: given $f'(x)$ as a product, find where $f$ increases or decreases, or how many turning points it has; given a cubic or quartic, find the local maximum or minimum value.

Example Multiple choice: reading the sign of the derivative

The function $y=f(x)$ has derivative $f'(x)=(x+1)(x-2)^2$ for all $x\in\mathbb{R}$ . On which interval is the function decreasing? A. $(-\infty, -1)$ . B. $(-1, 2)$ . C. $(2, +\infty)$ . D. $(-1, +\infty)$ .

Solution. Since $(x-2)^2\ge0$ , the sign of $f'(x)$ is the sign of $x+1$ (except at $x=2$ , where $f'=0$ ):

$f'(x)<0\iff x<-1,\qquad f'(x)\ge0\ \text{for}\ x\ge-1.$

So the function is decreasing on $(-\infty, -1)$ : the answer is A. The function has only one turning point, $x=-1$ ; at $x=2$ the derivative does not change sign.

In a true/false question, we compute the derivative and the turning values once for all four statements.

Example True/false: a cubic

Let $f(x)=-x^3+3x^2+9x-2$ . a) The function is increasing on $(-1, 3)$ . b) The function has a local maximum at $x=-1$ . c) The local maximum value of the function is $25$ . d) The line through the two turning points of the graph has gradient $8$ .

Solution. We compute the derivative and the values at its roots:

$f'(x)=-3x^2+6x+9=-3(x+1)(x-3);\qquad f(-1)=-7,\quad f(3)=25.$

The coefficient $-3<0$ , so $f'>0$ between the roots and $f'<0$ outside.

a) True: $f'(x)>0$ for $-1<x<3$ .

b) False: $f'$ changes from $-$ to $+$ at $x=-1$ , so this is a local minimum.

c) True: the local maximum is at $x=3$ , with value $f(3)=-27+27+27-2=25$ .

d) True: the turning points are $(-1, -7)$ and $(3, 25)$ , and the gradient is $\dfrac{25-(-7)}{3-(-1)}=\dfrac{32}{4}=8$ .

Key: T F T T.

Maximum and minimum values

Asymptotes

Note Remember

$x=x_0$ is a vertical asymptote if $\lim\limits_{x\to x_0^+}f(x)$ or $\lim\limits_{x\to x_0^-}f(x)$ is $\pm\infty$ .
$y=b$ is a horizontal asymptote if $\lim\limits_{x\to+\infty}f(x)=b$ or $\lim\limits_{x\to-\infty}f(x)=b$ .
$y=ax+b$ ( $a\ne0$ ) is an oblique asymptote if $\lim\limits_{x\to\pm\infty}\left[f(x)-(ax+b)\right]=0$ . For $y=\dfrac{x^2+x+1}{x-1}=x+2+\dfrac{3}{x-1}$ , the oblique asymptote is $y=x+2$ .
For $y=\dfrac{ax+b}{cx+d}$ ( $c\ne0$ , $ad-bc\ne0$ ): vertical asymptote $x=-\dfrac dc$ , horizontal asymptote $y=\dfrac ac$ .

A common trap: the denominator is $0$ at $x_0$ but so is the numerator; after cancelling, $x_0$ may no longer be a vertical asymptote.

Example Multiple choice: counting asymptotes

How many vertical and horizontal asymptotes does the graph of $y=\dfrac{x^2-1}{x^2-3x+2}$ have altogether? A. $2$ . B. $3$ . C. $1$ . D. $4$ .

Solution. We factorise the numerator and denominator and cancel for $x\ne1$ :

$y=\frac{(x-1)(x+1)}{(x-1)(x-2)}=\frac{x+1}{x-2}\quad(x\ne1).$

As $x\to1$ , $y\to-2$ (finite), so $x=1$ is not a vertical asymptote. As $x\to2^+$ , $y\to+\infty$ , so $x=2$ is a vertical asymptote. As $x\to\pm\infty$ , $y\to1$ , so $y=1$ is a horizontal asymptote.

So there are $2$ asymptotes: the answer is A. Option B wrongly counts $x=1$ .

Tables of variations and graph reading

From a table of variations or a graph we can read where the function increases or decreases, its turning points, its maximum and minimum, and the number of roots of $f(x)=k$ : the number of points where the graph meets the horizontal line $y=k$ .

Example True/false: reading a table of variations

The function $y=f(x)$ is continuous on $\mathbb{R}$ and has this table of variations:

$x$	$-\infty$		$-1$		$2$		$+\infty$
$f'(x)$		$+$	$0$	$-$	$0$	$+$
$f(x)$	$-\infty$	$\nearrow$	$3$	$\searrow$	$-2$	$\nearrow$	$+\infty$

a) The local maximum value of the function is $3$ . b) The function is increasing on $(-1, +\infty)$ . c) The equation $f(x)=1$ has three distinct roots. d) The equation $f(x)=3$ has three distinct roots.

Solution. a) True: $f'$ changes from $+$ to $-$ at $x=-1$ and the local maximum value is $3$ .

b) False: on $(-1, 2)$ the function is decreasing.

c) True: since $-2<1<3$ , the line $y=1$ meets the graph once on each of $(-\infty, -1)$ , $(-1, 2)$ , $(2, +\infty)$ .

d) False: the line $y=3$ only touches the graph at the local maximum $x=-1$ and crosses it once more on $(2, +\infty)$ , giving $2$ roots.

Key: T F T F.

To identify a cubic $y=ax^3+bx^2+cx+d$ from its graph, check in turn: if the right-hand branch rises, $a>0$ ; the graph meets the $y$ -axis at $(0, d)$ ; the $x$ -coordinates of the turning points are the roots of $y'$ .

The graph of a cubic

Example Multiple choice: identifying a graph

The curve in the figure above is the graph of which function? A. $y=x^3-6x^2+9x-1$ . B. $y=-x^3+6x^2-9x+1$ . C. $y=x^3-6x^2+9x+1$ . D. $y=x^3+6x^2+9x-1$ .

Solution. We eliminate options using features read from the figure. The right-hand branch rises, so $a>0$ : rule out B. The graph meets the $y$ -axis at $(0, -1)$ , so $d=-1$ : rule out C. For D, $y'=3(x+1)(x+3)$ has negative roots, which contradicts the figure (turning points at $x=1$ and $x=3$ ): rule out D.

Check A: $y'=3(x-1)(x-3)$ , $y(1)=3$ , $y(3)=-1$ , matching the figure. The answer is A.

Key terms

English	Tiếng Việt
increasing / decreasing	đồng biến / nghịch biến
local maximum / local minimum point	điểm cực đại / cực tiểu
turning value (local extreme value)	giá trị cực trị
maximum / minimum value	giá trị lớn nhất / nhỏ nhất
vertical / horizontal / oblique asymptote	tiệm cận đứng / ngang / xiên
table of variations	bảng biến thiên
repeated root of even multiplicity	nghiệm bội chẵn
gradient	hệ số góc
closed interval / open interval	đoạn / khoảng

Exercises

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1 Multiple choice

How many turning points does $y=x^4-2x^2+1$ have? A. $3$ . B. $1$ . C. $2$ . D. $0$ .

What you need and how to start

What you need: a turning point is a root of $y'$ where $y'$ changes sign.

Steps:

Factorise $y'=4x^3-4x$ .
Check the sign change at each root.

Show solution

$y'=4x(x-1)(x+1)$ has three simple roots $-1$ , $0$ , $1$ , and $y'$ changes sign at each of them.

Answer. A.

Exercise 2 Multiple choice

The vertical and horizontal asymptotes of the graph of $y=\dfrac{3x-1}{x-2}$ are A. $x=2$ and $y=3$ . B. $x=-2$ and $y=3$ . C. $x=2$ and $y=\dfrac12$ . D. $x=3$ and $y=2$ .

What you need and how to start

What you need: $y=\dfrac{ax+b}{cx+d}$ has vertical asymptote $x=-\dfrac dc$ and horizontal asymptote $y=\dfrac ac$ .

Steps:

Find the root of the denominator and check that the numerator is not $0$ there.
Take the ratio of the coefficients of $x$ .

Show solution

The denominator is $0$ at $x=2$ and the numerator is $5\ne0$ there, so $x=2$ is a vertical asymptote. $\lim\limits_{x\to\pm\infty}y=3$ . Option C takes the ratio of the constant terms $\dfrac{-1}{-2}$ .

Answer. A.

Exercise 3 Short answer

Let $M$ , $m$ be the maximum and minimum values of $y=x+\dfrac4x$ on $[1, 5]$ . Find $M+m$ .

What you need and how to start

What you need: compare the values at the roots of $y'$ in $(1, 5)$ and at the two endpoints.

Steps:

Solve $y'=1-\dfrac{4}{x^2}=0$ .
Compute $y(1)$ , $y(2)$ , $y(5)$ .

Show solution

$y'=0\iff x=2$ (reject $x=-2$ ). $y(1)=5$ , $y(2)=4$ , $y(5)=5+0.8=5.8$ . So $M=5.8$ and $m=4$ .

Answer. $9.8$ .

Exercise 4 Written answer

Find the asymptotes of the graph of $y=\dfrac{x^2+x+1}{x-1}$ .

What you need and how to start

What you need: divide the numerator by the denominator to write $y=ax+b+\dfrac{r}{x-1}$ .

Steps:

Carry out the polynomial division.
Look at the limits as $x\to1^\pm$ and at $y-(ax+b)$ as $x\to\pm\infty$ .

Show solution

The division gives $x^2+x+1=(x-1)(x+2)+3$ , so $y=x+2+\dfrac{3}{x-1}$ . As $x\to1^+$ , $y\to+\infty$ , so $x=1$ is a vertical asymptote. As $x\to\pm\infty$ , $y-(x+2)=\dfrac{3}{x-1}\to0$ , so $y=x+2$ is an oblique asymptote. There is no horizontal asymptote, because $y\to\pm\infty$ .

Answer. Vertical asymptote $x=1$ , oblique asymptote $y=x+2$ .

Exercise 5 True/False

The function $y=f(x)$ has derivative $f'(x)=x(x-1)^2(x+2)$ for all $x\in\mathbb{R}$ .

a) The function has three turning points.

b) The function has a local minimum at $x=0$ .

c) The function is increasing on $(0, +\infty)$ .

d) $f(-1)>f(0)$ .

What you need and how to start

What you need: a repeated root of even multiplicity does not make $f'$ change sign.

Steps:

Make a sign table for $f'$ with roots $-2$ , $0$ , $1$ .
Judge each statement.

Show solution

The sign of $f'$ : positive on $(-\infty, -2)$ , negative on $(-2, 0)$ , positive on $(0, 1)$ and on $(1, +\infty)$ .

a) False: there is no sign change at $x=1$ ; there are only two turning points, $x=-2$ and $x=0$ .

b) True: $f'$ changes from $-$ to $+$ at $x=0$ .

c) True: $f'\ge0$ on $(0, +\infty)$ and is $0$ only at $x=1$ .

d) True: $f$ is decreasing on $(-2, 0)$ and $-1<0$ .

Answer. a) False; b) True; c) True; d) True.

Exercise 6 Short answer

The concentration of a drug in the blood $t$ hours after it is taken is $C(t)=\dfrac{5t}{t^2+4}$ (mg/l), $t\ge0$ . What is the highest concentration, in mg/l?

What you need and how to start

What you need: the quotient rule $\left(\dfrac uv\right)'=\dfrac{u'v-uv'}{v^2}$ ; the maximum on $[0, +\infty)$ .

Steps:

Compute $C'(t)$ and find its positive root.
Find the sign of $C'$ , noting that $C(0)=0$ and $C(t)\to0$ as $t\to+\infty$ .

Show solution

$C'(t)=\frac{5(t^2+4)-5t\cdot2t}{(t^2+4)^2}=\frac{5(4-t^2)}{(t^2+4)^2}.$ For $t\ge0$ , $C'>0$ when $t<2$ and $C'<0$ when $t>2$ . The highest concentration is $C(2)=\dfrac{10}{8}=1.25$ mg/l, after $2$ hours.

Answer. $1.25$ .

Exercise 7 Multiple choice

The function $y=f(x)$ is continuous on $\mathbb{R}$ with this table of variations:

$x$	$-\infty$		$0$		$2$		$+\infty$
$f'(x)$		$+$	$0$	$-$	$0$	$+$
$f(x)$	$-\infty$	$\nearrow$	$1$	$\searrow$	$-3$	$\nearrow$	$+\infty$

The number of roots of $2f(x)+6=0$ is A. $1$ . B. $2$ . C. $3$ . D. $0$ .

What you need and how to start

What you need: the number of roots of $f(x)=k$ is the number of points where the graph meets the line $y=k$ .

Steps:

Rewrite as $f(x)=-3$ .
Read the table: $-3$ is the local minimum value.

Show solution

$2f(x)+6=0\iff f(x)=-3$ . The line $y=-3$ touches the graph at the local minimum $x=2$ and crosses the left branch once on $(-\infty, 0)$ (since $f$ rises from $-\infty$ to $1$ ). It does not meet the graph on $(2, +\infty)$ , where $f>-3$ .

Answer. B.

Exercise 8 True/False

Let $y=\dfrac{x+1}{x-1}$ .

a) The domain of the function is $\mathbb{R}\setminus\{1\}$ .

b) The function is decreasing on $\mathbb{R}\setminus\{1\}$ .

c) The asymptotes of the graph meet at $I(1, 1)$ .

d) The maximum value of the function on $[2, 4]$ is $3$ .

What you need and how to start

What you need: $y=\dfrac{ax+b}{cx+d}$ has $y'=\dfrac{ad-bc}{(cx+d)^2}$ ; monotonicity is stated on each interval separately.

Steps:

Compute $y'$ .
Find the two asymptotes and compare $y(2)$ , $y(4)$ .

Show solution

$y'=\dfrac{-1-1}{(x-1)^2}=\dfrac{-2}{(x-1)^2}<0$ for every $x\ne1$ .

a) True: the denominator is $0$ only at $x=1$ .

b) False: the function is decreasing on each of $(-\infty, 1)$ and $(1, +\infty)$ , but $y(0)=-1<y(2)=3$ , so it is not decreasing on the whole union.

c) True: vertical asymptote $x=1$ , horizontal asymptote $y=1$ .

d) True: the function is decreasing on $[2, 4]$ , so the maximum is $y(2)=3$ .

Answer. a) True; b) False; c) True; d) True.

Exercise 9 Short answer

Find the minimum value of $y=(x^2-3)e^x$ on $[0, 2]$ . Give your answer to 1 decimal place.

What you need and how to start

What you need: $(uv)'=u'v+uv'$ and $(e^x)'=e^x$ .

Steps:

Compute $y'=(x^2+2x-3)e^x$ and solve $y'=0$ on $(0, 2)$ .
Compare with $y(0)$ , $y(2)$ .

Show solution

$y'=2xe^x+(x^2-3)e^x=(x-1)(x+3)e^x=0\iff x=1$ (reject $x=-3$ ). $y(0)=-3$ , $y(1)=-2e\approx-5.44$ , $y(2)=e^2\approx7.39$ . The minimum value is $-2e\approx-5.4$ .

Answer. $-5.4$ .

Exercise 10 Challenge

The graph of $y=ax^3+bx^2+cx+d$ has a local maximum at $(0, 2)$ and a local minimum at $(2, -2)$ . Find $a$ , $b$ , $c$ , $d$ .

What you need and how to start

What you need: at a turning point $x_0$ , $y'(x_0)=0$ ; turning points lie on the graph.

Steps:

Write four equations: $y(0)=2$ , $y'(0)=0$ , $y(2)=-2$ , $y'(2)=0$ .
Solve, then check the type of each turning point.

Show solution

$y'=3ax^2+2bx+c$ . From $y(0)=2$ and $y'(0)=0$ : $d=2$ , $c=0$ . From $y'(2)=0$ : $12a+4b=0$ , so $b=-3a$ . From $y(2)=-2$ : $8a+4b+2=-2$ , so $8a-12a=-4$ , giving $a=1$ , $b=-3$ .

Check. $y=x^3-3x^2+2$ , $y'=3x(x-2)$ changes from $+$ to $-$ at $0$ and from $-$ to $+$ at $2$ ; $y(0)=2$ , $y(2)=-2$ .

Answer. $a=1$ , $b=-3$ , $c=0$ , $d=2$ .

Mini test

Marked like the Vietnamese graduation exam: Part I 0.25 points per question; Part II up to 1 point per question (1, 2, 3, 4 correct statements give 0.1, 0.25, 0.5, 1 point); Part III 0.5 points per question.

Part I. Multiple choice

Question 1. The function

y=f(x)

has derivative

f'(x)=(x-1)(x+3)

. On which interval is it increasing?

A. $(1, +\infty)$ .
B. $(-3, 1)$ .
C. $(-\infty, 1)$ .
D. $(-3, +\infty)$ .

Question 2. How many vertical asymptotes does the graph of

y=\dfrac{x+2}{x^2-4}

have?

A. $1$ .
B. $2$ .
C. $0$ .
D. $3$ .

Question 3. The maximum value of

y=-x^3+3x+1

[0, 2]

A. $3$ .
B. $1$ .
C. $-1$ .
D. $2$ .

Question 4.

The function $y=f(x)$ has this table of variations:

$x$	$-\infty$		$-1$		$1$		$+\infty$
$f'(x)$		$+$	$0$	$-$	$0$	$+$
$f(x)$	$-\infty$	$\nearrow$	$4$	$\searrow$	$0$	$\nearrow$	$+\infty$

The local minimum value of the function is

A. $0$ .
B. $1$ .
C. $4$ .
D. $-1$ .

Part II. True or false

Question 1. Let

f(x)=x^4-2x^2+3

a) $f'(x)=4x^3-4x$ .	True	False
b) The function has three turning points.	True	False
c) The local maximum value of the function is $2$ .	True	False
d) The minimum value of the function on $\mathbb{R}$ is $2$ .	True	False

Part III. Short answer

Question 1. The two asymptotes of the graph of

y=\dfrac{2x+1}{x-3}

meet at

I(a, b)

. Find

a+b

Question 2. Let

M

m

be the maximum and minimum values of

y=x+\dfrac1x

\left[\dfrac12, 3\right]

. Find

M+m

, to 1 decimal place.

Show answers and explanations

Part I, question 1: A. $f'(x)>0\iff x<-3$ or $x>1$ . Option B is where the function decreases; C and D contain part of $(-3, 1)$ , where $f'<0$ .
Part I, question 2: A. For $x\ne-2$ , $y=\dfrac{1}{x-2}$ ; as $x\to-2$ , $y\to-\dfrac14$ , so $x=-2$ is not a vertical asymptote. Only $x=2$ remains. Option B forgets to cancel.
Part I, question 3: A. $y'=-3x^2+3=0\iff x=1$ (reject $-1$ ). $y(0)=1$ , $y(1)=3$ , $y(2)=-1$ . Option B is $y(0)$ ; option C is the minimum value.
Part I, question 4: A. $f'$ changes from $-$ to $+$ at $x=1$ , so the local minimum value is $f(1)=0$ . Option B is the local minimum point, C the local maximum value, D the local maximum point.
Part II, question 1: a) True; b) True; c) False; d) True. a) True. b) True: $f'=4x(x-1)(x+1)$ changes sign at all three roots. c) False: the local maximum is at $x=0$ with value $3$ ; $2$ is the local minimum value. d) True: $f(\pm1)=2$ and $f(x)-2=(x^2-1)^2\ge0$ .
Part III, question 1: 5. The vertical asymptote is $x=3$ and the horizontal asymptote is $y=2$ , so $I(3, 2)$ and $a+b=5$ .
Part III, question 2: 5.3. $y'=1-\dfrac{1}{x^2}=0\iff x=1$ . $y\left(\dfrac12\right)=2.5$ , $y(1)=2$ , $y(3)=\dfrac{10}{3}$ . So $M+m=\dfrac{10}{3}+2=\dfrac{16}{3}\approx5.3$ .