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MEF University
Math 115 Calculus
Fall 2018-19 Midterm Exam 25 pts 25 pts 25 pts 25 pts 100 pts

Date: March 16, 2019 Name :

Time: 14:00-16:00 Surname :
Student ID :
IMPORTANT: 1. Show all your work to get full credit. 2. Calculators are not allowed.

1 1
1. (6+6+6+7 points) Given the functions f (x) = √ , g(x) = cot( ), k(x) = 2x − 1
2x − 3 x
and 
1/x , x < 0
h(x) =
 3x , x ≥ 0,

a) find the domain for each of the functions f and g,

Answer: For f, we need 2x − 3 > 0 ⇐⇒ x > 3/2, so

Dom(f ) = (3/2, +∞).
For g, we need x 6= 0 and x1 6= nπ for any n ∈ Z. So,
1
Dom(g) = R − ({0} ∪ { : n ∈ Z})
nπ
b) write (f ◦ g)(x) and (g ◦ f )(x) explicitly,
√
Answer: (f ◦ g)(x) = f (x) = √ 1
and (g ◦ f )(x) = cot( 2x − 3).
2 cot(1/x)−3

c) calculate (k ◦ h)(3) and (k ◦ h)(−1),

Answer: (k ◦ h)(3) = k(h(3)) = k(9) = 17 and (k ◦ h)(−1) = k(h(−1)) = k(−1) =

−3.
d) write (h ◦ k)(x) explicitly.

Answer: 
1/k(x) , k(x) < 0
h(k(x)) =
 3k(x) , k(x) ≥ 0,

So, 
1/(2x − 1) , x < 1/2
h(k(x)) =
 3(2x − 1) , x ≥ 1/2.
2. (7+6+6+6 points) Evaluate the following limits if they exist.

4x15/2 + 5x + 1
a) lim
x→∞ 2x8 + x3 − 5

Answer:
4x15/2 + 5x + 1 x15/2 (4 + 5x−13/2 + x−15/2 ) −1/2 (4 + 5x
−13/2
+ x−15/2 )
lim = lim = lim x = 0.
x→∞ 2x8 + x3 − 5 x→∞ x8 (2 + x−5 − 5x−8 ) x→∞ (2 + x−5 − 5x−8 )

x3 + x − 1
b) lim
x→∞ 5x3 + x5 sin( 12 )
x

Answer:
x3 + x − 1 x3 (1 + x−2 − x−3 ) 1 + x−2 − x−3 1 1
lim 3 5 1 = lim 3 2 1 = lim 1 = = .
x→∞ 5x + x sin( 2 )
x
x→∞ x (5 + x sin( 2 ))
x
x→∞
5+
sin( 2 )
x 5 + 1 6
1/x2

sin(1/x2 ) sin x
As we know lim 2 = 1 because lim = 1.
x→∞ 1/x x→0 x
1
sin( π−x )
c) lim−
x→π cot(x)
Answer:We recall
1
−1 ≤ sin( ) ≤ 1,
π−x
then
1
1 sin( π−x ) −1
≤ ≤
cot x cot x cot x
and as known
lim cot x = −∞
x→π −
1
sin( π−x )
we get by Sandwich theorem lim− = 0 together with the facts
x→π cot(x)
lim ± cot1 x = 0.
x→π −

sin(x4 − 1)
d) lim
x→1 x3 − 1
Answer:
sin(x4 − 1) sin(x4 − 1) x4 − 1 sin(x4 − 1) (x − 1)(x + 1)(x2 + 1)
lim = lim = lim
x→1 x3 − 1 x→1 x4 − 1 x3 − 1 x→1 x4 − 1 (x − 1)(x2 + x + 1)
sin(x4 − 1) (x + 1)(x2 + 1) (1 + 1)(1 + 1) 4
= lim 4 2
= = .
x→1 x −1 (x + x + 1) 1+1+1 3
3. a) (8 points) Find the values of a and b so that the function
(
2x3 − ax + 4 if x < 1
f (x) =
x2 + bx if x ≥ 1

is differentiable everywhere.

Answer: f must be continuous at x = 1, so lim+ f (x) = lim− f (x) =⇒ 2 −

x→1 x→1
a + 4 = 1 + b =⇒ a + b = 5. Furhermore, for differentiability of f we need
0 0
lim+ f (x) = lim− f (x) =⇒ 6 − a = 2 + b =⇒ a + b = 4 which is not possible for
x→1 x→1
we have a contradiction a + b = 4 and a + b = 5. Hence, there are no values of a and
b so that f is differentiable at 1.
dy sin x cos x
b) (8 points) Find , where y = .
dx x
 
dy d sin 2x 4x cos 2x − 2 sin 2x
Answer: = = .
dx dx 2x 4x2
√
2 π
c) (9 points) Find the slope of the line tangent to the graph of xy + tan(xy ) = 1 +
√ 2
at the point (1, 2π ).

Answer: We take derivatives of both side with respect to x by implicit differ-

entiation as follows
0 0
1.y + xy + sec2 (xy 2 )(1.y 2 + x2y.y ) = 0.
0
We solve for y and find
−y − y 2 sec2 (xy 2 )
0
y = .
x + 2xy sec2 (xy 2 )
√
Now, we substitute x = 1 and y = π/2 and obtain the slope as
√ √ √
− π/2 − (π/4) sec2 (π/4)

dy  − π/2 − (π/4)2 − π/2 − π/2
= √ = √ = √ .
dx x=1,y=√π/2 1 + 2( π/4) sec2 (π/4) 1 + 2( π/4)2 1+ π
4. (15+10 points) Given the function f (x) = x3 − 3x2 − 9x + 3,

a) Find the absolute maximum of f on the closed interval [−2, 2].

0
Answer: f (x) = 3x2 − 6x − 9 = 3(x + 1)(x − 3) = 0 =⇒ the critical points
as x = −1 and x = 3, but we do not consider x = 3 because 3 ∈ / [−2, 2].
We consider f (−2) = 1, f (−1) = 8, f (2) = −19. Hence, the absolute maximum of f
on [−2, 2] is 8.
b) Show that there is a real number c in the interval [0, 1] such that f (c) = 0.

Answer: f (0) = 3 > 0 > −8 = f (1), so by Intermediate theorem since f is

continuous we conclude that there is a real number c such that f (c) = 0.