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
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
