Topic 5: Graph Sketching,
Reasoning & Limits
Dr J Frost (jfrost@tiffin.kingston.sch.uk)
Last modified: 12th June 2015
Slide Guidance
?

Any box with a ? can be clicked to reveal the answer (this
works particularly well with interactive whiteboards!).
Make sure you’re viewing the slides in slideshow mode.
For multiple choice questions (e.g. SMC), click your choice to
reveal the answer (try below!)
Question: The capital of Spain is:
A: London

B: Paris

C: Madrid

Topic 5 – Graph Sketching and Reasoning
Part 1: Limits
RECAP: Differentiation by 1st Principles
Suppose we add some tiny value, ℎ, to 𝑥. Then:
?
𝑥 + ℎ, 𝑥 + ℎ
𝑥, 𝑥?2
The “lim” bit means
“what this expression
approaches as h tends
towards 0”
2
δy
δx
?
?
?
?
The h disappears as h
tends towards 0.
A question I often get asked is, why couldn’t we have substituted in ℎ = 0 straight away?
𝟎
The problem is that we’d end up with 𝟎, which
? is known as an indeterminate form.
Algebra of Limits
lim 𝑓 𝑥 + 𝑔 𝑥
𝑥→𝑐
= 𝐥𝐢𝐦 𝒇(𝒙) + ?𝒍𝒊𝒎 𝒈(𝒙)
𝒍𝒊𝒎 𝒇 𝒙
𝑓 𝑥
lim
= 𝒙→𝒄 ?
𝑥→𝑐 𝑔 𝑥
𝒍𝒊𝒎 𝒈 𝒙
𝒙→𝒄
𝒙→𝒄
𝒙→𝒄
lim 𝑓 𝑥 𝑔 𝑥
𝑥→𝑐
lim 𝑎𝑓 𝑥
𝑥→𝑐
lim 𝑎 𝑓
𝑥→𝑐
𝑥
= 𝒍𝒊𝒎 𝒇 𝒙
𝒙→𝒄
?𝒍𝒊𝒎 𝒈 𝒙
𝒙→𝒄
= 𝒂 𝒍𝒊𝒎 ?𝒇(𝒙)
𝒙→𝒄
𝒍𝒊𝒎 𝒇 𝒙
= 𝒂 𝒙→𝒄 ?
𝑙𝑖𝑚𝑥→𝑐 𝑙𝑜𝑔 𝑓 𝑥 = 𝒍𝒐𝒈 𝒍𝒊𝒎?𝒙→𝒄 𝒇 𝒙
1
lim
=0 ?
𝑥→∞ 𝑥
Examples
Determine
Useful in MAT!
𝑥
lim
𝑥→∞ 2𝑥+1
Bro Hint: Can we use that last law of limits?
Common Sense Reasoning Way
As 𝑥 becomes large, 2𝑥 and 2𝑥 + 1 are
barely different in the context of a
70
70
division (e.g. 150 vs 151). Thus:
𝑥
𝑥
1
?
≈
=
2𝑥 + 1 2𝑥 2
Formal Way
𝑥
1
= lim
1
𝑥→∞ 2𝑥 + 1
𝑥→∞
2+𝑥
1
=
?1
lim 2 + 𝑥
𝑥→∞
1
1
=
=
1
2
2 + lim 𝑥
𝑥→∞
lim
Examples
Use your knowledge of the relative growth rate of various functions to find these limits,
or state that they are divergent.
3
2𝑥 − 3𝑥 2
?
lim 2
=
−
𝑥→∞ 4𝑥 − 𝑥 + 1
4
lim
𝑥→∞
𝑥𝑥
𝑥!
Divergent.
𝑥 𝑥 grows
more rapidly
than 𝑥!
?
ln 𝑥
lim
𝑥→∞ 𝑥
𝑥 100
lim 𝑥
𝑥→∞ 2
= 0?
=0
?
Exponential
function always
‘beat’ polynomials.
Indeterminate Forms
0
Earlier, we saw that 0 causes a problem from the perspective of evaluating a limit.
This is known as an indeterminate form which, loosely speaking, are expressions that
have no value, or can proved to be multiple different values!
These are all the 7 different indeterminate forms, which are dealt with in different
ways when evaluating limits…
0
0
∞
∞
0
0
0×∞
∞−∞
?
∞
1
0
∞
Dealing with 0/0 and ∞/∞
0
Occasionally, as with our differentiation example earlier, limits which would yield 0
or
±∞
∞
can sometimes be dealt with by expanding brackets/simplifying:
lim 𝑥
𝑥
∞
𝑥→∞
How
lim 2 =
=could this go wrong?
2
𝑥→∞ 𝑥
lim 𝑥
∞
𝑥→∞
𝑥
1
lim 2 = lim How
= 0 should we do it?
𝑥→∞ 𝑥
𝑥→∞ 𝑥
But there’s other times where we couldn’t do this:
sin 𝑥
lim
𝑥→0
𝑥

l’Hôpital’s Rule
0
∞
l’Hôpital’s Rule allows us to transform a limit involving 0 or ∞ into a new quotient*
which is potentially not an indeterminate form.
If lim 𝑓 𝑥 = lim 𝑔 𝑥 = 0 but
𝑥→𝑐
𝑥→𝑐
𝑓′ 𝑥
lim ′
𝑥→𝑐 𝑔 𝑥
exists, then
𝑓 𝑥
𝑓′ 𝑥
lim
= lim ′
𝑥→𝑐 𝑔 𝑥
𝑥→𝑐 𝑔 𝑥
Example
This just means that if when we
differentiate the numerator and
denominator, we no longer have an
indeterminate form.
sin 𝑥
𝐜𝐨𝐬 𝒙
lim
= 𝐥𝐢𝐦
=𝟏
?
𝑥→0
𝒙→𝟎
𝑥
𝟏
*Remember that ‘quotient’ is just a posh word for the result of a division. Hence why we have the ‘quotient rule’.
Test Your Understanding
Determine
𝑥−1
lim
𝑥→1 ln 𝑥
𝑥−1
1
lim
= lim
= lim?𝑥 = 1
𝑥→1 ln 𝑥
𝑥→1 1/𝑥
𝑥→1
Determine lim 𝑥 ln 𝑥
𝑥→0
1
ln 𝑥
lim 𝑥 ln 𝑥 = lim
= lim? 𝑥 = lim −𝑥 = 0
1
𝑥→0
𝑥→0 1/𝑥
𝑥→0
𝑥→0
− 2
𝑥
(WolframAlpha.com can show you the step-by-step proof for these! e.g.
http://www.wolframalpha.com/input/?i=lim%5Bx-%3E0%5D%28x+ln%28x%29%29
You have to usually pay a subscription for the ‘proof’ feature, but if you buy the app for tablet/phone for a less
than 2 quid, it has this feature without subscription!)
Transforming to 0/0 and ∞/∞
1
ln 𝑥
lim 𝑥 ln 𝑥 = lim
= lim 𝑥 = lim −𝑥 = 0
1
𝑥→0
𝑥→0 1/𝑥
𝑥→0
𝑥→0
− 2
𝑥
Indeterminate form:
0 × −∞
Indeterminate form:
?
−∞
?
∞
In that second question, we cleverly turned a product (which was indeterminate)
into a quotient (which was indeterminate), in order to apply l’Hopital’s
? Rule.
Can we apply similar tricks to other kinds of expressions, to get a quotient?
0×∞
00 or 1∞
lim 𝑓 𝑥 𝑔(𝑥)
𝑥→𝑐
lim 𝑓 𝑥
𝑥→𝑐
𝑔 𝑥
𝑔 𝑥
lim
1?
𝑥→𝑐
𝑓 𝑥
𝑒
ln lim 𝑓 𝑥 𝑔 𝑥
𝑥→𝑐
?
…As we used above
=⋯
Why do you think we did this?
The ln allows us to move the 𝑔 𝑥 out the power and therefore get a product.
This in turn gives us 0 × ∞ and thus we?can apply the first transformation.
Example
Determine lim 𝑥 𝑥
𝑥→0
lim 𝑥 𝑥
𝑥→0
=𝑒
ln lim 𝑥 𝑥
𝑥→0
lim ln 𝑥 𝑥
=𝑒
lim
= 𝑒 𝑥→0
= 𝑒0
=1
𝑥→0
𝑥 ln 𝑥
0
As we have 0?
, apply relevant transformation.
We can move log inside the limit.
?
Use law of logs.
?
We proved this earlier.
?
?
Bonus Question
Bernoulli’s Compound Interest Problem:
I have £1, which I want to invest for 1 year. I get annual interest of 100%.
I can get the 100% in one instalment, and therefore finish with £2.
Or I can split the 100% into two instalments of 50%, and get £1 × 1.52 = £2.25
Or into ten instalments of 10%, and get £1 × 1.110 = £2.59. And so on.
In the limit, how much do I get?
1
lim 1 + ?
𝑛→∞
𝑛
=
=
=
=
=
𝑛
1 𝑛
lim 1+𝑛
𝑒 𝑛→∞
1
lim 𝑛 ln 1+𝑛
𝑒 𝑛→∞
1
ln 1+𝑛
lim
1
Laws of logs.
?
Indeterminate 0 × (−∞) so re-express as quotient.
𝑛→∞
𝑛
𝑒
1
− 2
𝑛
1
1+𝑛
lim
1
𝑛→∞
− 2
𝑛
𝑒
𝑒1 = 𝑒
?
?
The above was 1∞ , so apply associated transformation.
lim
=𝑒
1
1
1+𝑛
𝑛→∞
?
Apply l’Hopital’s and simplify.
Wow. It’s Euler’s constant!
?
Topic 5 – Graph Sketching and Reasoning
Part 2: Sketching Fundamentals
Graph Features?
What happens either side
of undefined ?
values of 𝑥?
y-intercept?
?
y as 𝑥 →
? ∞?
y as 𝑥 →
? −∞?
Turning ?
Points?
Roots??
Asymptotes?
?
y=
An asymptote is a straight line that a curve
approaches at infinity.
1
+ 2(𝑥 + 2)2
𝑥+2
Graph Features?
You may not have covered the following terminology yet:
-1
𝑥 ∈ ℝ, 𝑥? ≠ 0
The domain of the function is the possible values of the input:
The range of the function is the possible values of the output:
f 𝑥 ∈ ℝ, 𝑓? 𝑥 ≥ −1
The roots of the function are the inputs such that the output is 0:
The roots are also known as ‘zeros’ of the function, but never as the ‘x-intercepts’!
𝑥 𝑓 𝑥 = 0}
The two main ways of sketching graphs
1
Thing about the various features previous discussed.
2
And/or consider the individual components of the function
separately, and think how they combine.
We’re multiplying these
two individual functions
together.
𝑦 = 𝑥 sin 𝑥
What happens as 𝑥
increases?
What happens at the
peaks and troughs of
the sin graph?
Try sketching it!
𝑦 = 𝑥 sin 𝑥
?
Start say with 𝑦 = sin(𝑥). Whenever sin(𝑥) = 1, then
𝑥 sin 𝑥 = 𝑥. And whenever sin(𝑥) = −1, then
𝑥 sin(𝑥) = −𝑥. Notice also that when 𝑥 is negative,
multiplying sin(𝑥) by 𝑥 causes the graph to be flipped
on the 𝑦-axis.
1
𝑦=𝑥+
𝑥
Think about what
happens when we add
the graphs 𝑦 = 𝑥 and
1
𝑦=𝑥
𝒚=𝒙+
𝟏
𝒙
?
Notice the asymptote 𝑦 = 𝑥.
Non-vertical/horizontal
asymptotes are known as
oblique asymptotes.
𝑦 = sin 𝑥 /𝑥
𝒚 = 𝒔𝒊𝒏 𝒙 /𝒙
?
1
This is similar to the last, except we’re using 𝑦 = 𝑥 and 𝑦 =
1
− 𝑥 to work out the peaks and the troughs.
sin 𝑥
𝑛→0 𝑥
We can use our knowledge of limits to work out lim
!
𝑦 = 𝑥𝑥
Turning point:
𝑥
𝑦 = 𝑥 𝑥 = 𝑒 ln 𝑥 = 𝑒 𝑥 ln 𝑥
𝑑𝑦
= 1 + ln 𝑥 𝑒 𝑥 ln 𝑥 = 1 + ln 𝑥 𝑥 𝑥
𝑑𝑥
1 + ln 𝑥 𝑥 𝑥 = 0
1
1 + ln 𝑥 = 0 ∴ 𝑥 =
𝑒
lim 𝑥 𝑥 = 1
𝑥→0
Alternatively you could have used implicit
differentiation (C4).
?
𝟏 𝟏
,
𝒆 𝒆
?
𝟏
𝒆
When 𝑥 > 1, clearly 𝑥 𝑥 grows very rapidly
(faster than an exponential function).
Why is 𝒙𝒙 not well defined when 𝒙 < 𝟎?
Consider −
1
2
−
1
2
= −2
1
2
= −2
?
1
1 −3
−
3
1
3
This would happen whenever we have a fraction with an even denominator. However
= −3 3 = − 3,
which is a real number. We have a lack of continuity, as we could pick two fractions infinitely close together, one
which has an odd denominator, and the other even.
Values to a power
On the same axis, sketch 𝑦 = 𝑥 3 and 𝑦 = 𝑥 5 .
?
It’s easy to forget that for values
in the range −1 < 𝑥 < 1, the
higher the power we raise it to,
the smaller it becomes.
Exercise caution!
Values to a power
Draw a graph of 𝑦 = sin(𝑥). On the same axis, draw 𝑦 = sin2 𝑥
𝒚 = 𝒔𝒊𝒏 𝒙
𝒚 = 𝒔𝒊𝒏𝟐 𝒙
Notice that: (a) a squared value is always positive and
(b) when 0 < sin 𝑥 < 1, we’ll find that sin2 (𝑥) < sin 𝑥 .
3D graphs
Sketch 𝒛 = 𝒙𝟐 + 𝒚𝟐 (where 𝑥 and 𝑦 are on a horizontal plane and 𝑧 points upwards)
?
Explanation:
We know that 𝑥 2 + 𝑦 2 = 𝑘 is the equation of a circle (where 𝑘 is a constant).
Thus for a fixed value of 𝑧 we have a circle?(with radius 𝑧, thus 𝑧 must be positive).
As 𝑧 increases, the radius increases, and we get gradually get larger circles.
Composite functions
Sketch 𝑦 = sin 𝑥 2
?
As 𝑥 increases, 𝑥 2 increases more
rapidly. We’re effectively ‘speeding up’
across the sin graph, so the oscillation
period gradually decreases.
Note also that since 𝑓 −𝑥 = 𝑓(𝑥), we
have symmetry across the 𝑥-axis.
Composite functions
Sketch 𝑦 = sin
𝑥
?
As 𝑥 increases, 𝑥 increases less
rapidly. The input to the sin function
increases less rapidly, so we move
across the sin graph more slowly.
Note also that the domain is 𝑥 ≥ 0.
Reciprocal functions
Sketch 𝑦 =
1
sin 𝑥
Here’s a sketch of 𝑦 = sin 𝑥. What will
its reciprocal look like?
𝟏
𝒚=
𝒔𝒊𝒏 𝒙
1
0
leads to asymptotes.
1
1
1
= 1 and = −1, so the graphs
−1
touch for 𝑥 = 90°, 270°, …
Functions transforms
Sketch 𝑦 =
2
𝑥−1
1
Hint: If we start with 𝑓 𝑥 = 𝑥, how
could we transform this to get the
desired function?
𝑓 𝑥
1
𝑥
=
Click to view
transformation
1
𝑓 𝑥?
−1
=
1
𝑥−1
Click to view
transformation
-2
2𝑓 𝑥?− 1
=
2
𝑥−1
Putting it all together…
Because of multiplication by
2−𝑥 , peaks gradually
become shallower.
Because of 𝑠𝑖𝑛2 (and 2−𝑥 > 0
for all 𝑥), always positive.
The 𝑥 2 inside the 𝑠𝑖𝑛 causes
the oscillation period to
decrease.
We could have
eliminated this choice
by trying 𝑥 = 0.
A

[Source: Oxford MAT 2007]
B
C
D

A harder one
𝑥
−
1
𝑦2 =
𝑥+1
For this graph, it might be helpful to think about:
1. How do deal with the y2.
2. The asymptotes (both horizontal and vertical).
3. The domain of x (determine this once you’ve
dealt with the y2).
4. Roots.
𝑦 2 = (𝑥 − 1)/(𝑥 + 1)
Not defined for
-1<x<1.
As x becomes larger, the +1
and -1 has increasingly little
effect, so y = 1 for large x.
Repeated above and
below x-axis because
we have y = √...
Topic 5 – Graph Sketching and Reasoning
Part 3: Reasoning about Solutions
Polynomials
A polynomial expression is of the form:
𝑎0 + 𝑎1 𝑥 + 𝑎2 𝑥 2 + 𝑎3 𝑥 3 + ⋯
where 𝑎𝑖 are constants (which may be 0). Note that the powers of the variable must be
1
positive (or 0). 𝑥 2 + 𝑥 is not a polynomial expression as there’s a power of −2.
The degree of a polynomial is the highest power.
A polynomial of
degree 4.
𝑦4 − 𝑦 + 1
Names of polynomials:
Order 2
Quadratic
Order 3
Cubic
Order 4
Quartic
Order 5
Quintic
?
?
?
Polynomials
The range of a cubic is −∞ to ∞, i.e.
the entirety of ℝ.
The range of a quadratic however is
finite, as it has a maximum/minimum
(depending on whether the coefficient
of 𝑥 2 is negative or positive)
Can we generalise this to polynomials of any degree?
Polynomials
Polynomials of odd degree will always
have a range which spans the whole of
the real numbers.
Polynomials of even degree will always
have a finite range with a minimum or
maximum.
It goes ‘uphill’ if the coefficient of the
highest-power term is positive.
It will be ‘valley’ shaped if the coefficient
of the highest-power term is positive.
Number of Roots
What can we therefore say about the potential number of roots?
(where the polynomial has degree 𝑛)
Odd Degree
Even Degree
Minimum
Roots:
?1
0?
Maximum
Roots:
?𝑛
𝑛?
Number of Roots
We can shift a graph up and down by
changing the constant term.
i.e. the 𝑘 in 𝑦 = ⋯ + 𝑘
Click to Start Animation
Number of Distinct Roots:
01 2 3 4
Number of Roots
Question:
a) Sketch 𝑦 = 𝑥 4 − 6𝑥 2 + 9
b) For what values of 𝑏 does the equation y = 𝑥 4 − 6𝑥 2 + 𝑏 have the following
number of distinct roots (i) 0, (ii) 1, (iii) 2, (iv) 3, (v) 4.
[Source: STEP 1 (2012)]
2
2
a) By factorising, 𝑦 = 𝑥 − 3 . This is a
quartic, where 𝑦 is always positive, and has
repeated roots at 𝑥 = ± 3:
?
b)
By changing 𝑏, we shift the
graph up and down. Then we
can see that:
? 𝑏>9
i) 0 roots: When
ii) 1 root: Not possible.
?
iii) 2 roots: When b =
9 or 𝑏 < 0 ?
iv) 3 roots: 𝑏 =?0
v) 4 roots: 0 <?𝑏 < 9
Repeated factors/roots
Sketch: 𝑦 = 𝑥 2 (𝑥 − 1)
?
In general:
• When (𝑥 − 𝑎) appears once as a factor:
Line crosses 𝒙-axis at 𝒙 = 𝒂
• When 𝑥 − 𝑎 2 is a repeated factor:
Line touches 𝒙-axis?at 𝒙 = 𝒂
• When 𝑥 − 𝑎 3 is a doubly-repeated:
Point of inflection ?on 𝒙-axis at 𝒙 = 𝒂
Repeated factors/roots
Here’s a challenge!
Sketch: 𝑦 = 2𝑥 2 𝑥 − 1 𝑥 + 1
3
Inflection point
Touches axis
Crosses axis
?
The polynomial is of degree 6. So there’s up to 5 turning points. 2 are ‘used up’ by the inflection point. There’s 1
turning point at the origin. So that leaves 2 turning points left.
We can therefore deduce one turning point appears somewhere in −1 < 𝑥 < 0 and the other in 0 < 𝑥 < 1.
Turning Points
Polynomials’ turning points oscillate between 𝑚𝑖𝑛 → 𝑚𝑎𝑥 → 𝑚𝑖𝑛 → 𝑚𝑎𝑥 →
⋯ or 𝑚𝑎𝑥 → 𝑚𝑖𝑛 → 𝑚𝑎𝑥 → ⋯
𝒚= 𝒙−𝟏 𝒙−𝟐 𝒙−𝟑 𝟐 𝒙+𝟏
max
max
min
min
For a polynomial of order 𝑛, the maximum number of turning points is:
𝒏 −? 𝟏
Points of Inflection
However, when we have a point of inflection, then two of the turning points
effectively ‘conflate’ into one.
It’s a bit like having a max point immediately followed by a min (or vice versa)
(In fact we can have more than 2 turning points ‘conflate into one’. Consider 𝑦 = 𝑥 4 . A
quartic can have up to 3 stationary point, but this graph only has 1!)
Points of Inflection
A point of inflection is where the curve changes from curving downwards to
curving upwards (or vice versa).
It may or may not be a stationary point: (depending on whether 𝑓 ′ 𝑥 = 0)
Stationary point of inflection
Non-stationary point of inflection
Known as a ‘saddle-point’
Known as a ‘non-stationary point of inflection’
Strategies for determining number of solutions
METHOD 1: Reason about the graph
As we already have done:
• If it’s a polynomial, is the degree even or odd?
• If we have a constant that can be changed, consider the graph shifting up and down.
• We may have to find the turning points (by differentiation or completing the square)
METHOD 2: Factorise (when possible!)
e.g. 𝑥 3 − 𝑥 2 − 𝑥 + 1 = 0
This cubic conveniently factorises to:
𝑥2 𝑥 − 1 − 1 𝑥 − 1 = 𝑥2 − 1 𝑥 − 1
= 𝑥 + 1? 𝑥 − 1 2
We can see it has three solutions (two of them equal).
Look out for the difference of two squares!!!
METHOD 3: Consider the discriminant
Remember that if we have a quadratic 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, there are real solutions if:
𝑏 2 − 4𝑎𝑐 ≥ 0
?
Example
[Source: Oxford MAT 2009]
A

B
C
D

Example
• By differentiating to find the turning points: 12𝑥 3 − 48𝑥 2 + 36𝑥 = 0
𝑥 3 − 4𝑥 2 + 3𝑥 = 0
𝑥 𝑥 2 − 4𝑥 + 3 = 0
𝑥 𝑥−1 𝑥−3 = 0
So the turning points occur at 𝑥 = 0, 1, 3. Then considering the graph of the quartic:
If the x-axis is anywhere in the horizontal trip between the
maximum and the greater of the two minimums
(whichever it is), we’ll have four solutions because the line
will cross the axis 4 times. The y-values of the turning
points are 𝑘, 5 + 𝑘 and 𝑘 − 27 respectively. So 5 + 𝑘 > 0
so the maximum is above the x-axis, and 𝑘 < 0 so that the
greater of the two minimums occurs below the x-axis.
Example
[Source: Oxford MAT 2009]
A

B

C
D

Example
•
•
•
•
Spot when you can use the difference of two squares
Make use of the discriminant.
Thus 𝑥 4 − 𝑥 − 𝑐 2 = 0
𝑥2 + 𝑥 − 𝑐 𝑥2 − 𝑥 + 𝑐 = 0
1
2
If 𝑥 + 𝑥 − 𝑐 = 0, and using the discriminant on the first, 𝑐 ≥ − .
• Using the discriminant on the second 𝑐 ≤
1
4
4