1. [5 marks] Let
ax2
f (x) = ,
bx + cx3
where a, b and c are nonzero constants. Find the polynomials s(x) and l(x) such that
s(x) best approximates f (x) for small values of x, and l(x) best approximates f (x) for
large values of x.
For small x the term bx + cx3 is well approximated by bx and so f (x) ≈ ab x. For large x
a
the term bx + cx3 is well approximated by cx3 and so we have f (x) ≈ cx . This is a good
approximation, and full marks were given for it, but it’s true that it isn’t a polynomial.
The other accepted answer is f (x) ≈ 0, being the best polynomial approximation of f (x).
2. [5 marks] Find all the values of c such that
2
x + 2 if x ≤ c
f (x) =
4x − 1 if x > c
is continuous.
Both branches are polynomials and continuous functions. The only possible point of
discontinuity is at x = c. We therefore require c2 + 2 = 4c − 1. Solving gives c = 1 and
c = 3.
3. [5 marks] Let f (x) = 2x2 + 3x − 1. Use a definition of the derivative to find f ′ (0). No
credit will be given for solutions using differentiation rules, but you can use those to check
your answer.
We compute
2(x + h)2 + 3(x + h) − 1 − 2x2 − 3x + 1
f ′ (x) = lim
h→0 h
2
4xh + 2h + 3h
= lim
h→0 h
= lim 4x + 2h + 3
h→0
= 4x + 3
and so f ′ (0) = 3.
4. [5 marks] Find the slope of the tangent line to the curve
√
x−7
y=√
x+7
at x = 9.
First we compute the derivative using quotient rule to see
1 √ √
′
√
2 x
( x + 7) − 2√1 x ( x − 7)
y = √ 2 .
( x + 7)
, At x = 9 we have that the slope of the tangent line is
1
6
(10+ 4) 7
= .
100 300
5. [5 marks] Find the slope of the tangent line to the curve y = xx at x = e2 .
First, we rewrite y = ex log (x) . (Alternatively, take the logarithm
x logof(x)both sides and differ-
′ x
entiate implicitly.) Next differentiate to get y = log x + x e = (log x + 1)xx . So
2
the slope of the tangent line at x = e2 is 3e2e .
6. [5 marks] Find the equation of the tangent line to the curve
2
x2 + y 2 = 2x2 + 2y 2 − x
at the point 0, − 21 . Your answer should be in the form y = mx + b.
Let’s differentiate implicitly in y:
d d 2
x2 + y 2 = 2x2 + 2y 2 − x
dy dy
2x + 2yy = 2(2x2 + 2y 2 − x)(4x + 4yy ′ − 1).
′
We substitute x = 0 and y = − 21 and solve for y ′ :
1
−y ′ = 2 (−2y ′ − 1)
2
y = 2y ′ + 1
′
y ′ = −1.
So, our tangent line equation will take the form y = −x − b. It remains to find b by
substituting our point: − 21 = b. All together we have y = −x − 12 .
7. [5 marks] Use the degree 2 Taylor approximation to f (x) = cos(x) about x = 0 to
approximate cos 51 .
2
We use a second order Taylor approximation about x = 0: T2 (x) = f (0)+f ′ (0)x+f ′′ (0) x2! .
First we compute f (0) = cos(0) = 1 and f ′ (0) = − sin 0 = 0 and f ′′ (0) = − cos490 = −1 so
x2 1
that cos x ≈ T2 (x) = 1 − 2 . We therefore make the approximation cos 5 ≈ 50 .
2
8. [5 marks] Let f (x) = e−x +2x . Find all local extrema, and indicate clearly if each is a
local maximum or a local minimum.
2
We first find the derivative, f ′ (x) = e−x +2x (2 − 2x), to look for critical points. This
function exists everywhere so we solve f ′ (x) = 0 which yields only x = 1 since ex ̸= 0 for
any x. Our critical point x = 1 is a local maximum. There are several ways to see that
it is, in fact, a maximum:
2
ax2
f (x) = ,
bx + cx3
where a, b and c are nonzero constants. Find the polynomials s(x) and l(x) such that
s(x) best approximates f (x) for small values of x, and l(x) best approximates f (x) for
large values of x.
For small x the term bx + cx3 is well approximated by bx and so f (x) ≈ ab x. For large x
a
the term bx + cx3 is well approximated by cx3 and so we have f (x) ≈ cx . This is a good
approximation, and full marks were given for it, but it’s true that it isn’t a polynomial.
The other accepted answer is f (x) ≈ 0, being the best polynomial approximation of f (x).
2. [5 marks] Find all the values of c such that
2
x + 2 if x ≤ c
f (x) =
4x − 1 if x > c
is continuous.
Both branches are polynomials and continuous functions. The only possible point of
discontinuity is at x = c. We therefore require c2 + 2 = 4c − 1. Solving gives c = 1 and
c = 3.
3. [5 marks] Let f (x) = 2x2 + 3x − 1. Use a definition of the derivative to find f ′ (0). No
credit will be given for solutions using differentiation rules, but you can use those to check
your answer.
We compute
2(x + h)2 + 3(x + h) − 1 − 2x2 − 3x + 1
f ′ (x) = lim
h→0 h
2
4xh + 2h + 3h
= lim
h→0 h
= lim 4x + 2h + 3
h→0
= 4x + 3
and so f ′ (0) = 3.
4. [5 marks] Find the slope of the tangent line to the curve
√
x−7
y=√
x+7
at x = 9.
First we compute the derivative using quotient rule to see
1 √ √
′
√
2 x
( x + 7) − 2√1 x ( x − 7)
y = √ 2 .
( x + 7)
, At x = 9 we have that the slope of the tangent line is
1
6
(10+ 4) 7
= .
100 300
5. [5 marks] Find the slope of the tangent line to the curve y = xx at x = e2 .
First, we rewrite y = ex log (x) . (Alternatively, take the logarithm
x logof(x)both sides and differ-
′ x
entiate implicitly.) Next differentiate to get y = log x + x e = (log x + 1)xx . So
2
the slope of the tangent line at x = e2 is 3e2e .
6. [5 marks] Find the equation of the tangent line to the curve
2
x2 + y 2 = 2x2 + 2y 2 − x
at the point 0, − 21 . Your answer should be in the form y = mx + b.
Let’s differentiate implicitly in y:
d d 2
x2 + y 2 = 2x2 + 2y 2 − x
dy dy
2x + 2yy = 2(2x2 + 2y 2 − x)(4x + 4yy ′ − 1).
′
We substitute x = 0 and y = − 21 and solve for y ′ :
1
−y ′ = 2 (−2y ′ − 1)
2
y = 2y ′ + 1
′
y ′ = −1.
So, our tangent line equation will take the form y = −x − b. It remains to find b by
substituting our point: − 21 = b. All together we have y = −x − 12 .
7. [5 marks] Use the degree 2 Taylor approximation to f (x) = cos(x) about x = 0 to
approximate cos 51 .
2
We use a second order Taylor approximation about x = 0: T2 (x) = f (0)+f ′ (0)x+f ′′ (0) x2! .
First we compute f (0) = cos(0) = 1 and f ′ (0) = − sin 0 = 0 and f ′′ (0) = − cos490 = −1 so
x2 1
that cos x ≈ T2 (x) = 1 − 2 . We therefore make the approximation cos 5 ≈ 50 .
2
8. [5 marks] Let f (x) = e−x +2x . Find all local extrema, and indicate clearly if each is a
local maximum or a local minimum.
2
We first find the derivative, f ′ (x) = e−x +2x (2 − 2x), to look for critical points. This
function exists everywhere so we solve f ′ (x) = 0 which yields only x = 1 since ex ̸= 0 for
any x. Our critical point x = 1 is a local maximum. There are several ways to see that
it is, in fact, a maximum:
2
