CALCULUS EXAM SOLUTIONS

Detailed Solutions To A4I Exams

Irakli Diasamidze
2023-2024
Contents

1 Year 2023-2024 2
1.1 Midterm 1, Version 1 . . . . . . . . . . . . . . . . . . . . . . . 2

1
Chapter 1

Year 2023-2024

1.1 Midterm 1, Version 1

Problem 1
Let L be a line passing at a point (a, b). Suppose that L is parallel to the
tangent line of the graph of y = f (x) at the point (x0 , f (x0 )). Write equation
4
of L if f (x) = − x1/4 + 3x1/3 ; x0 = 1; (a, b) = (−1, 1). (5 points)

In general, the equation of a line passing at a point (a, b) is given by

y = k(x − a) + b with k, the slope, some real number. Since we want the
line L to pass at the point (a, b) = (−1, 1) and be parallel to the tangent line
of the graph of y = f (x) at the point (x0 , f (x0 )), its equation shall be given
as y = k(x − a) + b = k(x − (−1)) + 1 = k(x + 1) + 1, where we also have
k = f ′ (x0 ) = f ′ (1).
To calculate the latter, we first differentiate f as follows.
 ′  1 ′  1 ′
′ 4
f (x) = − 1/4 + 3x 1/3
= −4 x− 4 + 3 x 3 =
x
 
1 5 1 2 5 2
= −4 · − x− 4 + 3 · x− 3 = x − 4 + x− 3
4 3

5 2
Meaning f ′ (1) = 1− 4 + 1− 3 = 2. Hence the line L is given by y =
2(x + 1) + 1 = 2x + 3.

2
Problem 2
d
Calculate dx |x|, x ̸= 0. Show that f (x) = |x| does not have derivative at
x = 0. (4 points)

Let x ̸= 0. We then have two cases x > 0 or x < 0.

d
For x > 0, we have the following for dx |x|.
d |x + h| − |x| (x + h) − x
|x| = lim = lim =
dx h→0 h h→0 h
h
= lim = lim 1 = 1
h→0 h h→0

d
For x < 0, we have the following for dx
|x|.
d |x + h| − |x| −(x + h) − (−x)
|x| = lim = lim =
dx h→0 h h→0 h
−h
= lim = lim [−1] = −1
h→0 h h→0

Finally, we prove that the derivative of f (x) = |x| at x = 0 does not exist
by showing that its one-sided limits of limh→0 f (0+h)−f
h
(0)
exist and are not
equal. Note that 0 + h = h and f (0) = 0.
f (h) − f (0) |h| − 0 h
lim+ = lim+ = lim+ = lim 1 = 1
h→0 h h→0 h h→0 h h→0
f (h) − f (0) |h| − 0 −h
lim = lim− = lim− = lim− [−1] = −1
h→0− h h→0 h h→0 h h→0

f (0+h)−f (0)
Since 1 ̸= −1, the limit limh→0 h
, i.e., the derivative of f (x) at
x = 0, does not exist.

Problem 3
Show that the function x5 = x3 − 3x + 5 has a root in the interval (1, 2). (5
points)

3
 We assume that the author meant the equation x5 = x3 − 3x + 5 rather
than the function.
Consider the function f (x) = x5 − x3 + 3x − 5. Note that f (1) = 15 − 13 +
3(1) − 5 = −2 and f (2) = 25 − 23 + 3(2) − 5 = 25. Since f (1) = −2 < 0 <
25 = f (2) and f , is continuous on the closed interval [1, 2], by intermediate
value theorem, there exists x ∈ (1, 2) with f (x) = 0, i.e., x5 − x3 + 3x − 5 = 0
and x5 = x3 − 3x + 5 for some x ∈ (1, 2), meaning a root in the interval (1, 2)
indeed exists.

Problem 4
Verify that the point (−1, 3) is on the curve of the implicit function x2 y 2 = 9
and find the value of the slope of the tangent line at the point. (4 points)

We verify that the point (−1, 3) indeed lies on the given curve by plugging
x = −1, y = 3 into the equation and seeing if the obtained equation is true.
If x = −1 and y = 3, then x2 y 2 = (−1)2 32 = 9 indeed holds.
Via implicit differentiation, an equation with y ′ in it is yielded. To obtain
the slope of the tangent line at the point (−1, 3), we solve for y ′ with x = −1,
y = 3 plugged in.

x2 y 2 = 9
(xy)2 = 9

′
(xy)2 = (9)′
2xy · (xy)′ = 0
2xy(xy ′ + y) = 0
2(−1)(3)(−y ′ + 3) = 0
−y ′ + 3 = 0
y′ = 3