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MAT237Y1
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Dan Dolderman
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Mathematics

MAT237Y1

Dan Dolderman

Fall

Description

Calc 1
Fri 10/24/08
5.3 - Graphing junk
Example: Let’s graph another polynomial before going on to bigger and better things. Let
2 2 3
g(x) = x (x − 1) (x + 1) . Graph this sucker.
Roots: 0,1,-1
y-intercept: g(0) = 0. So, the y-intercept is at the origin.
lim g(x) = +∞
x→∞
x→−∞ g(x) = −∞
f (x) = x(x − 1)(x + 1) (7x − x − 2)
√ √
1 − 57 1 + 57
Set this equal to zero and x = 0,1,−1, ≈ −.47, ≈ .61. These are our critical
14 14
points. As for the second derivative,
f (x) = 2(x + 1)(21x − 6x − 14x + 2x + 1)
And that’s zero when x = −1, −.71, −.22, .36, .86. Time for the second derivative test and time
to look for changes in concavity!
′′
f (0) = 2 so we have a min
′′
f (1) = 16 this is a min
f (−1) = 0 boo we’ll need to do more work for this!
√ !
1 − 57
f ′′ = −1.5 max!
14
√ !
1 + 57
f ′′ = −4.7 again, a max!
14
Now if we go ahead and do the ﬁrst derivative test at x = −1, then we’ll get pos to pos, so it’s
neither a max nor a min. As for concavity,
−−−−−−− +++++++ −−−−−−− +++++++ −−−−−−− +++++++
−1 −.71 −.22 .36 .86
1 0.4
y
0.2
–1.5 –1 –0.5 0 0.5 1 1.5
x
–0.2
–0.4
Now let’s go back and think about this the “precalc” way: Notice that at each of the roots, -1 had
a multiplicity of 3, so it crossed the x-axis, it was tangent to the x-axis, and there was an inﬂection
point there. Also, both 0 and 1 had multiplicities of 2 and so both touched, but did not cross the
x-axis, while being tangent to the x-axis at those points, plus the concavity didn’t change!
Okay, now it’s time to deal with graphing rational functions. Recall that a rational function is the
quotient of two polynomials. The idea is that whenever you have a rational function, before you
even think about graphing it, make sure you cancel out any factors. Note where those holes are! It
will make things so much easier if you cancel out the like terms (just remember to write down
those punctures when you graph it!).
Five Steps in order to Sketch a Rational Function
(1) Solve for the x-intercepts (and y-intercept if you’d like. The book says do this so that you
know where the function crosses the y-axis. It’s not necessary, though). Notice that when
you solve for the x-intercepts, you only need to worry about the numerator because the
denominator will never make the function 0!
(2) Take the denominator and set it equal to 0. These will be your vertical asymptotes (this is
why we got rid of the common factors before so we don’t have to worry about if they are VA
or punctures).
(3) solve
lim f(x)
x→∞
and
lim f(x)
x→−∞
to see what happens. If you get a ﬁnite number, this means there is a horizontal asymptote.
If you get ±∞, see oblique asymptotes.
2 (4) calculate the sign of f(x) in between each root and vertical asymptote (like when we ﬁnd if
f(x) is increasing, we plug in test values and see if it’s positive). This isn’t necessary, but
the more information you have about a function, the better. What could be easier than
plugging in a couple numbers into your original function to see if it gives you a negative or
positive number out?
(5) Finally, do all that stuﬀ from 5.1 and 5.2 (increasing, decreasing, ccu, ccd, inﬂection points,
relative extrema, stationary points).
Whew! Once you get all this stuﬀ calculated, you can ﬁnally sketch the graph of a rational
function! Okay, that was a lot to digest. Let’s do a few examples of sketching graphs.
Example: Sketch the graph of
2
2x
f(x) = x − 1
First, ﬁnd out if there are any common factors.
2 2
2x 2x
x − 1 = (x + 1)(x − 1)
Nope! So, let’s begin.
(1) x-intercepts:
2
f(x) = 2x = 0
x − 1
means 2x = 0 or x = 0.
y-intercepts: 2
2(0) 0
f(0) = 2 = = 0
0 − 1 −1
(2) Vertical Asymptotes:
x − 1 = (x + 1)(x − 1) = 0
so x = 1,−1.
(3) Horizontal Asymptotes:
2
lim 2x = 2
x→∞ x − 1
can use L’Hopital’s or end behavior here.
2x2 2(−x)2
lim 2 = lim 2 = 2
x→−∞ x − 1 x→∞ (−x) − 1
So, one HA at y = 2.
(4) The signs of f(x)
+++++++ −1 −−−−−−− 0 −−−−−−− 1 +++++++
−2 −1/2 1/2 2
3 (5) all that other stuﬀ.
4x(x − 1) − 2x (2x) −4x
f (x) = =
(x − 1) 2 (x − 1) 2
′
where is f (x) zero and undeﬁned? at x = 0,−1,1. Check signs.
+++++++ +++++++ −−−−−−− −−−−−−−
−2 −1 0 1 2
−1/2 1/2
So increasing on (−∞,−1) ∪ (−1,0] while decreasing [0,1) ∪ (1,∞). Our stationary (critical)
point is x = 0. Using ﬁrst derivative test, going from pos to neg, so relative max. Don’t
bother with x = −1,1 because we found out they were already VA. How can they be relative
max’s or min’s if they shoot up to ∞ or −∞? Remember: VA’s can never be critical points!
What you’re looking at has to be in the domain of f. In this case, x = −1,1 are not in the
domain of f!
concavity:
2 2 2 2
f (x) = −4(x − 1) + (4x)2(x − 1)2x = 12x + 4
(x − 1) 4 (x − 1) 3
′′
f (x) can never be zero, so we need to check the signs for concavity for x = −1,1.
+++++++ −−−−−−− +++++++
−1 1
−2 0 2
CCU (−∞,−1) ∪ (1,∞) and CCD (−1,1). There are no inﬂection points bceause although
the concavity changes at 1 and -1, they are not originally in the domain. Remember this!!!
Inﬂection points HAVE to be in the original domain!! Otherwise, they’re not points; they’re
asymptotes.
Now that we have ﬁnally ﬁnished all these steps, we can sketch the curve.
4
y
2
0
–4 –3 –2 –1 1 2 3 4
x
–2
–4
4 Example: Sketch the following function:
3
f(x) = x + 1
x − 1
Does it reduce? nope.
(1) intercepts.
x-intercepts: x + 1 = 0. Only gives x = −1.
y-intercept:
0 + 1
f(0) = = −1
0 − 1
3
(2) VA: x − 1 = 0 gives x = 1.
(3) HA:
3
lim x + 1 = 1
x→∞ x − 1
and
x + 1
lim = 1
x→−∞ x − 1
(4) the signs of f
+++++++ −1 −−−−−−− 1 +++++++
−2 0 2
(5) the other stuﬀ
2 3 3 2 2
′ 3x (x − 1) − (x + 1)(3x ) −6x
f (x) = 3 2 = 3 2
(x − 1) (x − 1)
x = 0 gives us that possibility for a relative extremum, so let’s check signs (don’t forget to
check x = 1 too!)
−−−−−−− −−−−−−− −−−−−−−
−1 0 1/2 1 2
So f(x) is decreasing (−∞,1) ∪ (1,∞).
5 Concavity:
3 2 2 3 2 3 3 3
′′ −12x(x − 1) + 6x 2(x − 1)(3x ) −12x(x − 1 − 3x ) 12x(2x + 1)
f (x) = (x − 1) 4 = (x − 1) 3 = (x − 1) 3
possible inﬂection pts: x = 0 and x = − √3 (don’t forget to check x = 1!!!)
2
−−−−−−−− 3 +++++++ −−−−−−− +++++++
−1 −1/ 2 −1/2 0 1/2 1 2
1 1 3
CCU (− 3 ,0) ∪ (1,∞) and CCD (−∞,− √3 ) ∪ (0,1). Both x = 0 and x = −1/ 2 are
2 2
inﬂection points because there is a change in concavity.
Time to graph.
4
y
2
–4 –3 –2 –1 0 1 2 3 4
x
–2
–4
6 A little blurb about Oblique Asymptotes:
Recall earlier how when we looked at HA’s, they all went to some number (they can also go to
zero, but the particular examples i had didn’t do that). Well, something else can happen when we
take the limit of a rational function. These are where Oblique Asymptotes occur. They replace
what would be a horizontal asymptote. Yes, the limit goes to inﬁnity, but how? It can follow
2
along a path like y = x, or perhaps like y = x , or it could do some other crazy thing; who knows!
Essentially it means this: reduce the rational function like you would from an improper fraction to
a mixed fraction (using, what else, but polynomial long division!). Here it is in action:
x3
Example: Sketch the curve of f(x) = x +1 .
Are there any common factors? Nope
(1) intercepts:
x = 0 means x = 0. Also, f(0) = 0.
(2) VA: none
(3) now instead of HA, we’re looking for oblique asymptotes.
divide x + 1 into x . Yay for long division!
x
x + 1 x3
3
− x − x
− x
So
x3 −x
= x +
x + 1 x + 1
Notice how we were able to reduce that down to there. This means that the oblique
asymptote is the line y = x (the “whole” part of that mixed fraction). So, instead of there
being a HA at, say y = 3, where the function tails oﬀ to 3 as x → ∞, this “HA” is now the
line y = x.
How do we know when we have oblique asymptotes? When the degree of the numerator is
bigger than the degree of the denominator. Notice all of the earlier examples were when the
numerator was the same degree as the denominator.
(4) The signs of f(x).
−−−−−−− +++++++
−1 0 1
(5) the rest
3x (x + 1) − x (2x) x (x + 3)
f′(x) = =

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