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Practice exam solution and Lecture content for big class and small class

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Course
MATH 100

Institution
University Of British Columbia (UBC )

There is 2 practice exam which is similar to the final exam in term of format and difficulties. Big class and small class content include what will be tough in the lecture and some practice problems from a downloadable pdf book

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Uploaded on December 19, 2022
Number of pages 10
Written in 2022/2023
Type Class notes
Professor(s) Mark mac lean, usman muhammad, anthony wachs
Contains All classes

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1. [5 marks] Trace the graph of f (x) = 5x3 − 5x4 on the axes below, using the dotted
curves.
The blue curve below is the correct graph of f (x).

x+2
2. [5 marks] Find all the horizontal and vertical asymptotes of f (x) = √ .
4x2 + 3x + 2
This function has no vertical asymptotes. To see this note that 4x2 + 3x + 2 > 0 for all x.
1
This function has two horizontal asymptotes: y = 2
and y = − 12 . To find these, compute
1 1
lim f (x) = and lim f (x) = − .
x→∞ 2 x→−∞ 2
3. [5 marks] Find all the values of m such that
3
6x − 2m if x ≤ −1
f (x) =
2x2 + 5m if x > −1
is continuous.

, We note that both branches are polynomials and hence continuous on their domain. The
only place where f (x) might be discontinuous is at the point x = −1. We therefore
require that 6(−1)3 − 2m = 2(−1)2 + 5m. Solving for m yields m = − 87 .

4. [5 marks] Let f (x) = x5 . Use a definition of the derivative to find f ′ (x). No credit will be
given for solutions using differentiation rules, but you can use those to check your answer.
We compute

f (x + h) − f (x)
f ′ (x) = lim
h→0 h
5 5
−
= lim x+h x
h→0 h
5x − 5(x + h) 1
= lim ·
h→0 x(x + h) h
−5h 1
= lim ·
h→0 x(x + h) h
−5
= lim
h→0 x(x + h)
5
= − 2.
x

5. [5 marks] Suppose the function f (x) is defined and continuous on all real numbers, and
that
f (5 + h) − 3
lim = −5.
h→0 h
Given this information, we can find the equation of the tangent line to the curve y = f (x)
at a particular point. What is the point, and what is the slope of the tangent line to the
curve y = f (x) at that point?
From the above we notice that we are considering f ′ (5) with f (5) = 3 filled in. So, the
point in question is (5, 3) and the slope is −5.

2

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