Topic 5: Graph Sketching, Reasoning & Limits

Dr J Frost (jfrost@tiffin.kingston.sch.uk)

Last modified: 12th June 2015

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ζPart 1: LimitsTopic 5 – Graph Sketching and Reasoning

RECAP: Differentiation by 1st Principles

(𝑥 ,𝑥2 )

(𝑥+h , (𝑥+h )2 )

Suppose we add some tiny value, , to . Then:

The “lim” bit means “what this expression approaches as h tends towards 0”

The h disappears as h tends towards 0.

δxδy

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A question I often get asked is, why couldn’t we have substituted in straight away? The problem is that we’d end up with , which is known as an indeterminate form.?

Algebra of Limits

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Examples

Determine

Common Sense Reasoning Way

As becomes large, and are barely different in the context of a division (e.g. vs ). Thus:

Formal Way

lim𝑥→∞

𝑥2 𝑥+1= lim𝑥→∞

1

2+ 1𝑥

Bro Hint: Can we use that last law of limits?

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Useful in MAT!

ExamplesUse your knowledge of the relative growth rate of various functions to find these limits, or state that they are divergent.

lim𝑥→∞

2 𝑥−3 𝑥2

4 𝑥2−𝑥+1¿− 34

lim𝑥→∞

ln𝑥𝑥 ¿0

lim𝑥→∞

𝑥𝑥

𝑥 !

Divergent. grows more rapidly than

lim𝑥→∞

𝑥100

2𝑥 Exponential function always ‘beat’ polynomials.

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Indeterminate Forms

Earlier, we saw that 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…

00∞∞ 0×∞ ∞−∞

00 1∞ ∞ 0

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Dealing with and Occasionally, as with our differentiation example earlier, limits which would yield or can sometimes be dealt with by expanding brackets/simplifying:

lim𝑥→∞

𝑥𝑥2

=lim𝑥→∞

𝑥

lim𝑥→∞

𝑥2=∞∞

lim𝑥→∞

𝑥𝑥2

= lim𝑥→∞

1𝑥=0

How could this go wrong?

How should we do it?

But there’s other times where we couldn’t do this:

lim𝑥→ 0

sin (𝑥 )𝑥

l’Hôpital’s Rulel’Hôpital’s Rule allows us to transform a limit involving or into a new quotient* which is potentially not an indeterminate form.

If but exists, then

This just means that if when we differentiate the numerator and denominator, we no longer have an indeterminate form.

Example

lim𝑥→0

sin (𝑥 )𝑥 =𝐥𝐢𝐦

𝒙→𝟎

𝐜𝐨𝐬 (𝒙 )𝟏 =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

lim𝑥→ 1

𝑥−1ln 𝑥 =lim

𝑥→ 1

11/𝑥=lim

𝑥→ 1𝑥=1

Determine

lim𝑥→ 0

𝑥 ln𝑥= lim𝑥→ 0

ln 𝑥1/𝑥=lim

𝑥→ 0

1𝑥

− 1𝑥2

= lim𝑥→0

(− 𝑥 )=0

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(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%29You 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 and

lim𝑥→0

𝑥 ln𝑥= lim𝑥→0

ln 𝑥1/𝑥=lim

𝑥→ 0

1𝑥

− 1𝑥2

= lim𝑥→0

(− 𝑥 )=0

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?

Indeterminate form: Indeterminate form:

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lim𝑥→𝑐

𝑓 (𝑥 )𝑔(𝑥 )0×∞ lim𝑥→𝑐

𝑔 (𝑥 )

( 1𝑓 (𝑥 ) ) …As we used above

or lim𝑥→𝑐

𝑓 (𝑥 )𝑔 (𝑥 )

𝑒ln lim

𝑥→ 𝑐𝑓 (𝑥 )𝑔 ( 𝑥)

=…

Why do you think we did this?The ln allows us to move the out the power and therefore get a product. This in turn gives us and thus we can apply the first transformation.

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Example

Determine

lim𝑥→0

𝑥𝑥As we have , apply relevant transformation.

We can move log inside the limit.

Use law of logs.

We proved this earlier.

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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 Or into ten instalments of 10%, and get . And so on.In the limit, how much do I get?

lim𝑛→∞ (1+ 1𝑛 )

𝑛The above was , so apply associated transformation.

Laws of logs.

Indeterminate so re-express as quotient.

Apply l’Hopital’s and simplify.

Wow. It’s Euler’s constant!

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ζPart 2: Sketching FundamentalsTopic 5 – Graph Sketching and Reasoning

y= 1𝑥+2 +2¿

Turning Points?

Asymptotes?

y as ?

y as ?

y-intercept?

Roots?

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An asymptote is a straight line that a curve approaches at infinity.

Graph Features?What happens either side of undefined values of ??

Graph Features?

You may not have covered the following terminology yet:

-1

The domain of the function is the possible values of the input:

The range of the function is the possible values of the output:

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’!

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Thing about the various features previous discussed.

And/or consider the individual components of the function separately, and think how they combine.

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The two main ways of sketching graphs

𝑦=𝑥 sin (𝑥 )

𝒚=𝒙

𝒚=𝒔𝒊𝒏𝒙We’re multiplying these two individual functions together.

What happens as increases?

What happens at the peaks and troughs of the sin graph?

Try sketching it!

Start say with . Whenever , then . And whenever , then . Notice also that when is negative, multiplying by causes the graph to be flipped on the -axis.

𝑦=𝑥 sin (𝑥 )

𝒚=𝒙

𝒚=−𝒙

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𝑦=𝑥+1𝑥

𝒚=𝒙

𝒚=𝒙+𝟏𝒙

Notice the asymptote .Non-vertical/horizontal asymptotes are known as oblique asymptotes.

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Think about what happens when we add the graphs and

This is similar to the last, except we’re using and to work out the peaks and the troughs. We can use our knowledge of limits to work out !

𝑦=sin (𝑥 )/𝑥

𝒚= 𝟏𝒙

𝒚=−𝟏𝒙

𝒚=𝒔𝒊𝒏 (𝒙 )/𝒙 ?

When , clearly grows very rapidly (faster than an exponential function).

𝑦=𝑥𝑥

𝒚=𝒙𝒙

Turning point:

Alternatively you could have used implicit differentiation (C4).

(𝟏𝒆 ,(𝟏𝒆 )𝟏𝒆 )?

Why is not well defined when ?Consider This would happen whenever we have a fraction with an even denominator. However , 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.

lim𝑥→0

𝑥𝑥=1

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Values to a powerOn the same axis, sketch and .

It’s easy to forget that for values in the range , the higher the power we raise it to, the smaller it becomes.Exercise caution!

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Values to a powerDraw a graph of . On the same axis, draw

Notice that: (a) a squared value is always positive and (b) when , we’ll find that .

𝒚=𝒔𝒊𝒏𝟐 (𝒙 )𝒚=𝒔𝒊𝒏 (𝒙 )

3D graphsSketch (where and are on a horizontal plane and points upwards)

We know that is the equation of a circle (where is a constant).Thus for a fixed value of we have a circle (with radius , thus must be positive).As increases, the radius increases, and we get gradually get larger circles.

Explanation:

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Composite functionsSketch

As increases, increases more rapidly. We’re effectively ‘speeding up’ across the sin graph, so the oscillation period gradually decreases.Note also that since , we have symmetry across the -axis.

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Composite functionsSketch

As increases, 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 .

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Reciprocal functionsSketch

and , so the graphs touch for leads to asymptotes.

Here’s a sketch of . What will its reciprocal look like?

𝒚=𝟏

𝒔𝒊𝒏 𝒙

𝒚=𝒔𝒊𝒏 (𝒙

)

Functions transformsSketch Hint: If we start with , how could we

transform this to get the desired function?

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1

𝑓 (𝑥 ) 1𝑥¿

Click to view transformation

𝑓 (𝑥−1 ) 1𝑥−1¿

Click to view transformation

2 𝑓 (𝑥−1 ) 2𝑥−1¿

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A B C D

[Source: Oxford MAT 2007]

Putting it all together…Because of multiplication by , peaks gradually become shallower.

Because of (and for all ), always positive.

The inside the causes the oscillation period to decrease.

We could have eliminated this choice by trying .

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.

A harder one

𝑦 2=𝑥−1𝑥+1

Not defined for-1<x<1.

Repeated above and below x-axis because we have y = √...As x becomes larger, the +1

and -1 has increasingly little effect, so y = 1 for large x.

𝑦 2=(𝑥−1)/(𝑥+1)

ζPart 3: Reasoning about SolutionsTopic 5 – Graph Sketching and Reasoning

Polynomials

A polynomial expression is of the form:

where are constants (which may be 0). Note that the powers of the variable must be positive (or 0). is not a polynomial expression as there’s a power of .

The degree of a polynomial is the highest power.

Names of polynomials:

A polynomial of degree 4.

Order 2 Quadratic

Order 3 Cubic

Order 4 Quartic

Order 5 Quintic

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Polynomials

The range of a cubic is to , i.e. the entirety of .

The range of a quadratic however is finite, as it has a maximum/minimum (depending on whether the coefficient of 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.

It goes ‘uphill’ if the coefficient of the highest-power term is positive.

Polynomials of even degree will always have a finite range with a minimum or maximum.

It will be ‘valley’ shaped if the coefficient of the highest-power term is positive.

Number of RootsWhat can we therefore say about the potential number of roots? (where the polynomial has degree )

Minimum Roots:

Maximum Roots:

Odd Degree Even Degree

1 0𝑛 𝑛

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Number of Distinct Roots:

0 1 2 3 4

Number of Roots

Click to Start Animation

0 1 2 3 4

We can shift a graph up and down by changing the constant term.i.e. the in

Question: a) Sketch b) For what values of does the equation have the following number of distinct roots

(i) 0, (ii) 1, (iii) 2, (iv) 3, (v) 4.[Source: STEP 1 (2012)]

By factorising, . This is a quartic, where is always positive, and has repeated roots at :

By changing , we shift the graph up and down. Then we can see that:

i) 0 roots: When ii) 1 root: Not possible.iii) 2 roots: When iv) 3 roots: v) 4 roots:

Number of Roots

a) b)

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Repeated factors/roots

Sketch:

In general:• When appears once as a factor:

Line crosses -axis at • When is a repeated factor:

Line touches -axis at • When is a doubly-repeated:

Point of inflection on -axis at

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Repeated factors/rootsHere’s a challenge!

Sketch: 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 and the other in .

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Turning PointsPolynomials’ turning points oscillate between or

max

min

max

min

𝒚=(𝒙−𝟏 ) (𝒙−𝟐 ) (𝒙−𝟑 )𝟐 (𝒙+𝟏 )

For a polynomial of order , the maximum number of turning points is: ?

Points of InflectionHowever, 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 . A quartic can have up to 3 stationary point, but this graph only has 1!)

Points of InflectionA 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 )

Stationary point of inflectionKnown as a ‘saddle-point’

Non-stationary point of inflectionKnown as a ‘non-stationary point of inflection’

Strategies for determining number of solutions

METHOD 2: Factorise (when possible!)

e.g. This cubic conveniently factorises to:

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 , there are real solutions if:

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)

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A B C D

[Source: Oxford MAT 2009]

Example

• By differentiating to find the turning points:

So the turning points occur at . 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 and respectively. So so the maximum is above the x-axis, and so that the greater of the two minimums occurs below the x-axis.

Example

A B C D

Example[Source: Oxford MAT 2009]

• Spot when you can use the difference of two squares • Make use of the discriminant.• Thus • If , and using the discriminant on the first, . • Using the discriminant on the second

Example