MAC 2311-01
CALCULUS
I
TEST
I
SOLUTIONS
SUMMER
C
LAST NAME, NAME: True, Always
2017
TOTAL:
/100
June 5, 2017
5
, and the closed interval [0, 1] be given. Is there a value of c with 0 ≤ c ≤ 1 such that f (c) = 2? If you said
2x + 1
“Yes,” find the value of c. If you said “No,” explain why that is the case. Hint: The Intermediate Value Theorem.
1. Let f (x) =
The only discontinuity of f (x) is at x = − 21 which is not in the given interval [0, 1]. We have a continuous function at
5
= 2 to get x = 34 which is in the given
hand, and f (0) = 5 and f (1) = 35 . Since 53 ≤ 2 ≤ 5, we can solve f (x) =
2x + 1
interval [0, 1].
√
x2 + 4 − 2
2. Evaluate the limit without using tables. lim
.
x→0
x
√
x2 + 4 − 2
Let f (x) =
. When we substitute the limit value into the function, we see that it is of the form 00 . So, we
x
√
must rationalize the numerator of the function. That is, multiply top and bottom by x2 + 4 + 2. This will transform the
x2
function into f (x) = √
.
x( x2 + 4 + 2)
√
x2 + 4 − 2
x
√
x2 + 4 − 2
lim
x→0
x
√
√
x2 + 4 − 2
x2 + 4 + 2
=
· √
x
x2 + 4 + 2
2
x
x
=
= √
√
2
x(
x2 + 4 + 2
x + 4 + 2)
x
0
= lim √
= = 0.
x→0
x2 + 4 + 2 4
2
y
x
−6
−4
−2
2
4
−2
r
3. Evaluate the limit without using tables. lim
x→∞
3
3x + 5
.
6x − 8
r
Using the properties of limits, we get by simple substitution L =
4. Evaluate the limit without using tables. lim
x→∞
h√
i
x6 + 5x3 − x3 .
1
3
1
.
2
6
2
When we substitute the limit value into the function, we see that it is of the form ∞ − ∞. So we rationalize the numerator:
√
√
√
x6 + 5x3 + x3
3
3
6
3
6
3
x + 5x − x = ( x + 5x − x ) · √
x6 + 5x3 + x3
3
5x
5
=
q
= q
x3
1 + x53 + 1
1 + x53 + 1
lim
h√
x→∞
x6 + 5x3 − x3
5. Evaluate the limit without using tables. lim
x→∞
i
=
lim q
x→∞
5
1+
5
x3
+1
=
5
.
2
!
`n (2x)
.
`n (3x)
First, when we substitute the limit value into the function, we see that it is of the form
of the natural logarithmic function, we get `n (2x) = `n 2 + `n x. Hence, we have
`n 2
`n
x `n
`n 2 + `n x
x +1
lim
= lim
3
x→∞ `n 3 + `n x
x→∞
`n
x `n
`n x + 1
=
2
`n 2
`n x
x→∞ `n 3
`n x
lim
+1
+1
= 1.
y
1
x
−6
−4
−2
2
4
−1
−2
6. For what values of a and b, if any, is


2x − a if x < −3,







ax + 2b if −3 ≤ x ≤ 3,
f (x) = 






 b − 5x if 3 < x,
continuous at every x?
We need to find the left and right limits of f (x) at x = ±3.
lim f (x)
=
lim f (x)
=
lim f (x)
=
lim f (x)
=
x→−3−
x→−3+
x→3−
x→3+
lim (2x − a) = −6 − a
x→−3−
lim (ax + 2b) = −3a + 2b
x→−3+
lim (ax + 2b) = 3a + 2b
x→3−
lim (b − 5x) = b − 15
x→3+
6
∞
. Second using the properties
∞
3
We have to have lim− f (x) = lim+ f (x) and lim− f (x) = lim+ f (x). These yield a − b = 3 and 3a + b = −15, respectively.
x→−3
x→−3
Solving these gives a = −3, and b = −6.
x→3
x→3
y
−6
−4
x
−2
2
6
4
−5
−10
−15
−20
−25
7. Find the x-values (if any) at which f (x) is discontinuous. f (x) = exp
function e g(x) . Are the discontinuities removable?
The only possible problem might be x = 0. But since lim
x→0
sin x
, here exp(g(x)) means the natural exponential
x
sin x
= 1, we have a removable discontinuity at x = 0.
x
3
y
2
1
x
−3
−2
−1
8. Find the vertical asymptotes (if any) of the function f (x) =
2
1
3
2x
(ex
+ 1)(1 +
x2 )(1
− x2 )(x2 − π)
.
We have to solve (ex + 1)(1 + x2 )(1 − x2 )(x2 − π) = 0 in real numbers. It is easy to see that f (x) has vertical asymptotes at
√
x = ±1, ± π.
9. Given the graph of the function f (x) whose domain is (−∞, −5) ∪ (−5, 8], evaluate the limits if they exist. If the limit does not
exist, write DNE:
(a) lim+ f (x) = 2
(e) lim+ f (x) = DNE
(b) lim− f (x) = 0
(f) lim− f (x) = −6
(c) lim f (x) = DNE
(g) lim f (x) = DNE
(d) f (0) = −4
(h) f (8) = 2
x→0
x→0
x→0
x→8
x→8
x→8
4
y
10
8
6
4
2
−10 −8
−6
−4
−2−2
−4
−6
−8
−10
x
2
4
6
8
10