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Math 151 Section D
Midterm 1 Fall 2006
Solutions - Version 1
- [3 marks] Mark each statement T (True) or F (False):
F If f (x) < 0 for all x and limx→ 0 f (x) exists then limx→ 0 f (x) < 0.
F A function f is differentiable at c if its graph has a
tangent line at the point (c, f (c)).
F If limx→ 1 f (x) = ∞ and limx→ 1 g(x) = ∞ then limx→ 1 [f (x)/g(x)]
must be equal to 1.
F A function can have three different horizontal asymptotes.
T A function f is differentiable on an open interval (a, b)
if it is differentiable at every number in that interval.
T If f is differentiable at c, then f is continuous at c.
- (a) [1] Show by means of an example that limx→ 1 (f (x)·g(x)) may exist
even though neither limx→ 1 f (x) nor limx→ 1 g(x) exists. Justify
your answer.
Solution: One possible example is:
f (x) =
2 if x ≤ 1 1 2
if x > 1
and g(x) =
1 2
if x ≤ 1
2 if x > 1
Since
lim x→ 1 −^
f (x) = lim x→ 1 +^
g(x) = 2
and
lim x→ 1 +^
f (x) = lim x→ 1 −^
g(x) = 1/ 2
we see that neither limx→ 1 f (x) nor limx→ 1 g(x) exists. On the
other hand f (x) · g(x) = 1 for all x ∈ R.Therefore limx→ 1 (f (x) ·
g(x)) = 1.
(b) [1] Draw a graph of a function with a removable discontinuity.
Solution: See Figure 1.
1
2
3
4
y
–1 1 2 3 4 5 x
Figure 1: Removable Discontinuity
(c) [1] Give an example of a function f that is continuous for all real
numbers and such that f
′ is not defined at x = 3. Write a formula
and draw a graph of f.
Solution: One such a function is f (x) = |x − 3 |. See Figure 2.
1
2
3
4
y
–1 1 2 3 4 5 x
Figure 2: Continuous but Not Differentiable
- Evaluate the following limits. Justify your answers.
(a) [2] limx→π(x − π) cos
1
π−x
(b) [2] limx→ 5
25 −x^2
2 x^2 − 17 x+
Solution:
lim x→ 5
25 − x
2
2 x^2 − 17 x + 35
= lim x→ 5
(5 − x)(5 + x)
(x − 5)(2x − 7)
= lim x→ 5
−(5 + x)
2 x − 7
(c) [2] limx→∞
3 x^3 − 2 x^2 +2x− 1
4 − 2 x+5x^2 − 2 x^3
Solution:
lim x→∞
3 x
3 − 2 x
2
4 − 2 x + 5x^2 − 2 x^3
= lim x→∞
2 x
2 x^2
1 x^3 4 x^3
2 x^2
5 x
- (a) [2] State the Intermediate Value Theorem.
Solution: Suppose that f is continuous on the closed interval
[a, b] and let N be any number between f (a) and f (b), where
f (a) 6 = f (b). Then there exists a number c in (a, b) such that
f (c) = N.
(b) [4] Show that the equation x
3 − 4 x + 1 = 0 has three differ-
ent roots by calculating the values of the left-hand side at x =
− 3 , − 2 , − 1 , 0 , 1 , 2 , 3 and then applying the Intermediate Value
Theorem.
Solution: Let f (x) = x
3 − 4 x + 1. Since f is a polynomial we
conclude that f is continuous on R.
From
x − 3 − 2 − 1 0 1 2 3
f (x) − 14 1 4 1 − 2 1 16
we see that f (−3) < 0, f (−2) > 0, f (1) < 0, and f (2) > 0. By
the Intermediate Value Theorem, there are c 1 between −3 and
−2, c 2 between −2 and 1, and c 3 between 1 and 2 such that
f (c 1 ) = f (c 2 ) = f (c 3 ) = 0.
Therefore c 1 , c 2 , and c 3 are three different roots of the given equa-
tion.
- (a) [2] Define the derivative of a function f at a number a.
Solution: The derivative of a function f at a number a, denoted
by f
′ (a), is
f
′ (a) = lim h→ 0
f (a + h) − f (a)
h
if this limit exists.
(b) [4] Let f (x) =
2 − x. Use the definition of the derivative to find
f
′ (−2).
Solution:
f
′ (−2) = lim h→ 0
f (−2 + h) − f (−2)
h
= lim h→ 0
2 − (−2 + h) −
h
= lim h→ 0
4 − h − 2
h
4 − h + 2 √ 4 − h + 2
= lim h→ 0
4 − h − 4
h(
4 − h + 2)
= lim h→ 0
4 − h + 2
(c) [2] Find an equation of the tangent line to the curve y = f (x) = √ 2 − x at the point (− 2 , f (−2)).
Solution: From f (−2) =
2 − (−2) = 2 we get that an equation
of the tangent line is
y − 2 = −
· (x + 2).
- Let f (3) = 3, g(3) = 2, f
′ (3) = −1 and g
′ (3) = −2. Evaluate:
(a) [2] F
′ (3) if F (x) = e
x g(x) −
3
x
2 f (x).
Solution: From
F
′ (x) = e
x g(x) + e
x g
′ (x) −
x
− 1 / 3 f (x) −
3
x
2 f
′ (x)
