Answers to exam 2, version 2 — Math 1850-016 — Fall 03
f (x + h) − f (x)
1. For each of the following functions, evaluate
, and simplify
h
your result.
1
(a) f (x) = .
x
1
1
x−x−h
1
1
−
=
=−
h x+h x
hx(x + h)
x(x + h)
√
(b) f (x) = x.
√
√
1
x+h− x
x+h−x
= √
√ =√
√
h
h( x + h + x)
x+h+ x
2. (a) Find a formula for the family of linear functions of slope 3 and sketch
several members of this family.
A linear function whose graph has slope 3 must have slope-intercept equation
f (x) = 3x + b. This is a family of parallel lines, as pictured.
(b) Find a formula for the family of linear functions such that f (1) = 0 and sketch
several members of this family. (Use a new coordinate frame.)
A linear function whose graph passes thru (1, 0) has point-slope formula f (x) =
m(x − 1). This second family is pictured above, right.
(c) Which function belongs to both families? Identify it in each of your graphs.
The linear function common to these two families has equations f (x) = 3(x −
1) = 3x − 3. Its graph has slope 3 and passes thru (1, 0). This line is highlighted
in both graphs above.
3. If an object is thrown into the air with velocity 24 ft/s, its height in feet after
t seconds is given by y = 24t − 16t2 .
(a) Find the average velocity for the time period beginning when t = 1 and lasting
∆t seconds.
vave =
y(1 + ∆t) − y(1)
24 + 24∆t − 16 − 32∆t − 16∆t2 − 8
=
= −8 − 16∆t.
∆t
∆t
(b) Find the instantaneous velocity when t = 1.
The instantaneous velocity is the limit of vave as ∆t → 0. From part (a) we
find that this limit is −8.
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4. For the function f (whose graph is given) state the value of the given quantity,
if it exists. If it does not, explain why.
(a) lim f (x) = 1. (b) lim− f (x) does not exist since the functional values oscillate
x→0
x→1
wildly to the left of 1. (c) lim+ f (x) = 1. (d) lim f (x) does not exist since the
x→3
x→1
functional values become infinitely large near 3.
5. Evaluate lim x cos(2/x). Justify your answer in detail.
x→0
Since −1 ≤ cos(θ) ≤ 1 for every θ, we have that − |x| ≤ x cos(2/x) ≤ |x| for
every nonzero x. By the Squeeze Theorem we conclude that
lim − |x| = lim x cos(2/x) = lim |x| = 0.
x→0
x→0
x→0
6. Sketch the graph of an example of a function f that satisfies all of the given
conditions:
lim f (x) = 1.
x→1−
lim f (x) = −1.
lim f (x) = 2.
x→3
x→1+
f (3) undefined.
7. For each of the following, evaluate the limit if it exists. If it does not exists
explain why not.
√
x−2
−2
1
=
= .
x−4
−4
2
2
2
(x + h) − x
x2 + 2xh + h2 − x2
(b) lim
= lim
= lim 2x + h = 2x.
h→0
h→0
h→0
h
h
x4 − 1
(x − 1)(x + 1)(x2 + 1)
(c) lim
= lim
= lim (x + 1)(x2 + 1) = 4.
x→1 x − 1
x→1
x→1
x−1
(a) lim
x→0
2
8. Use the given graph to find a number δ such that |f (x) − 1| < 0.5 whenever
0 < |x − 2| < δ. Illustrate your solution. (You should probably trace the graph
into your blue book.)
3
2
1.5
1
0.5
1
2 2.2
3
If we draw the horizontal line y = 1.5 and find where it intersects the graph
near the point (2, 1), then trace from this intersection straight down to the x-axis
we find the point 2.2 (approximately). The graph always stays above the line
y = 0.5. So we may take δ to be 2.2 − 2, which is 0.2.
9. Graph the given function f . Determine whether or not it is continuous at
x = 3. Explain!
 2
x − 9
, if x 6= 3
f (x) =
x−3

6,
if x = 3
When f (x) 6= 3 we have f (x) = x + 3. Hence lim f (x) = 6, which is equal to
x→3
f (3). We conclude that the function is continuous at 3.
10. Find an equation of the tangent line to the curve y = 3x2 at the point where
x = 1. Sketch the curve and its tangent line.
We first find the tangent slope m.
3 + 6h + 3h2 − 3
y(1 + h) − y(1)
= lim
= lim 6 + 3h = 6.
h→0
h→0
h→0
h
h
m = lim
Hence the tangent line has point-slope equation y − 3 = 6(x − 1).
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