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A midterm exam for a calculus course, with various calculus-related questions covering limits, derivatives, integrals, and asymptotes. Students are expected to use the definition of derivatives, limits, and long division of polynomials to solve the problems.
Typology: Exams
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Midterm Exam Calculator Portion
Will the real ________________________ please stand up?
. Using the definition of the derivative, find (^) f '( ). x 6 points
0 0
lim lim h h
f x h f x (^) x h x
→ (^) h → h
0
lim h
x x h
x h x
→ h
0
lim h
x x h
x h x
→ h
0
lim h
h
x h x
→ h
0 2
lim h → ( x h 5)( x 5) ( x 5)
2.) Given
2 x k if x
kx if x f x. Find the value of k such that (^) f ( ) x is continuous
everywhere. 4 points
The two pieces of the function are polynomials, so they are continuous everywhere. The only concern is at the breaking
point of x =-1. For f to be continuous at -1, the left and right side limits must be equal so that the limit exists.
f ( 1)− = k ( 1)− − 3 = − k − 3 2
1 1
lim ( ) lim [( 1) ] 1 x x
f x k k →− +^ →−+
1 1
lim ( ) lim [ ( 1) 3] 3 x x
f x k k →− −^ →−−
Setting the left and right-side limits equal gives the following.
1 + k = − k − 3
2 k = − 4 → k = − 2
3.) Suppose that
x x
x
x x
x
x x
f x. 8 points
(a) Determine 2
lim ( ) x
f x →−+
. (c) Determine lim ( ) 0
f x x →
(b) Determine lim ( ) 2
f x x →
. (d) Determine 4
lim ( ) x
f x →−
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4.) Determine all vertical, horizontal, and slant asymptotes where they apply for 2 3 4 ( ). 5 1
x x f x x
Label the asymptotes. 4 points
Vertical asymptote at
x =
No horizontal asymptotes
Slant asymptote at
y = x + (Use long division of polynomials.)
5.) A spherical balloon is expanding. If the radius is increasing at the rate of 2 inches per minute, at what
rate is the volume increasing when the radius is 5 inches? 5 points
(^43)
3
V = π r
(^42) 3 3
dV dr r dt dt
π
(^42) 3 (5) (2) 3
dV
dt
π
(^42) 3 (5) (2) 3
dV
dt
π
3 3 200 in / m in 628.32in / m in
dV
dt
= π ≈
6.) Use logarithmic differentiation to find the derivative of
co s
x
cos ( )
x f x = x
co s ln ( ) ln x f x = x
1 1 '( ) ( sin )(ln ) (c o s ) ( )
f x x x x f x x
= − +
1 f '( x ) ( s in x )(ln x ) (c o s x ) f ( x ) x
= (^) − + ⋅
(^1) c o s '( ) ( s in ) (ln ) ( c o s ) x f x x x x x x
= − + ⋅
7.) Use the Squeeze Theorem to determine the following limit. 4 points
4 0 6
lim sin x
x → (^) x
6
1 sin 1 x
4 4 4 6
x x sin x x
4 4 4 0 0 6 0
lim lim sin lim x x x
x x x → → (^) x →
4 4 0 6 0 6
0 lim sin 0, which means lim sin must be 0. x x
x x → (^) x → x
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10.) The graph of y = f ( ) x is given below. From this graph, sketch a graph of y = f '( ) x on the
same set of axes. 4 points
11.) Determine coefficients a, b, and c so that the curve f(x) = ax
2
through the point (1,3) and line 4x + y = 8 is tangent to the curve at the point (2, 0).
Since f ( ) x passes through (1,3) and (2,0), both points must satisfy the function. This results in
two equations.
2 3 = a (1) + b (1) + c a + b + c = 3
2 0 = a (2) + b (2) + c 4 a + 2 b + c = 0
Since the line 4 x + y = 8 is tangent to the curve at (2,0), its slope must be the same as the
derivative of the function at x=2. (The slope of the line is -4.) This results in a third equation.
f '( ) x = 2 ax + b and at x = 2 with the slope of the tangent line equal to 4, the resulting equation is 4 a + b = −4.
So,
a + b + c = 3
4 a + 2 b + c = 0
4 a + b = − 4
Working with the first two equations, you can eliminate variable c. Multiply the first equation by
-1 and add the two equations together. This result in the new equation of 3 a + b = − 3. Working
with this new equation and third equation, you can eliminate another variable. [I’m choosing b.
You can multiply either equation by -1 and then add them together. This results in...
a = − 1
Now, work backwards. If a = − 1 , then b = 0 ,which means c = 4.