Fall 2011, Math 1A - Calculus
GSI: Shishir Agrawal
Discussions 102 and 103
Worksheet 4
Problem 1. Determine the domain and range of each of the following functions.
(a) f pxq ?x
(c) hpxq ?
3
(b) g pxq lnp2 3xq
1
1 4x2
Solution.
(a) Observe that f pxq is the inverse of the function x3 . The domain and range of x3 are both R, so the
domain and range of f pxq are also R.
(b) The expression lnp2 3xq is valid if and only if 2 3x ¡ 0, if and only if 2 ¡ 3x, if and only if x 2{3.
Thus the domain of g pxq is p8, 2{3q. The range of g pxq is all real numbers (why?).
(c) Observe that hpxq is valid whenever 1 4x2 ¡ 0 (why?). This implies that 4x2 1, so x2 1{4. Thus,
the domain of hpxq is all x such that 1{2 x 1{2. In other words, the interval p1{2, 1{2q.
To figure out the range of hpxq?
, observe that as x Ñ 1{2 and x Ñ 1{2 , hpxq Ñ 8. Also, hpxq
reaches a minimum value when 1 4x2 reaches a maximum value, which happens when x 0 (why?).
Since hp0q 1, this suggests that the range is the interval r1, 8q.
Problem 2. Determine whether or not each of the following functions has an inverse. If it does, find a
formula for the inverse.
(a) f pxq ex
(b) g pxq 2
?
9 2x
(c) hpxq lnpx3
1q
Solution.
(a) The function f pxq has no inverse (why?).
(b) ?
The function g pxq can be written as the composition of the functions 9 2x, which is one-to-one, and
x, which is also one-to-one. Thus g pxq is also one-to-one, so has an inverse.
?
To calculate its inverse, we set x 9 3y and then solve for y to get y p1{3qx2 3. Thus, the
inverse is given by g 1 pxq p1{3qx2 3 for all x ¥ 0 (why do we have to say for all x ¥ 0?).
(c) The function hpxq can be written as the composition of x3 1, which is one-to-one, and lnpxq, which is
also one-to-one. Thus hpxq has an inverse. To calculate the inverse, we set x lnpy 3 1q and solve for
y (do it!).
1
$
px
'
'
'
& x
Problem 3. Let
2q2
if x ¤ 2,
if 2 x ¤ 0,
if 0 x ¤ 1, and
if x ¡ 1.
2
f p xq '
lnpxq
'
'
%
x1
Determine where f pxq is continuous. Then write a definition for any function g pxq which is continuous on
all of R and such that g pxq f pxq on p8, 2s Y p1, 8q.
Solution. The function f pxq is continuous for all x 2, 0 (calculate left- and right-handed limits at
and 1 to verify this).
The function
$
2
'
&px 2q if x ¤ 2,
g p xq 0
if 2 x ¤ 1
'
%x 1
if x ¡ 1.
2, 0
is continuous on all of R and agrees with f pxq on p8, 2sYp1, 8q. There are many other possible solutions
as well.
Problem 4. Evaluate the following limits.
x2
Ñ1 1 x
a
(b) lim 3x 9x2 2x
xÑ8
(a) lim
x
(c) lim sinpxq
x
Ñ8
(d) lim
x
(e)
Ñ2
f pxq f p2q
where f pxq x4
x2
x2 2x
xÑ8 4x2 x
lim
?
x2
3
1
2 x
xÑ4 4x x2
(f) lim
Solution.
x2
(a) Observe that as x Ñ 1 , px 2q Ñ 1, and 1 x becomes a small negative number. Thus lim
xÑ1 1 x
8.
(b) Compare to problem 1 on quiz 3 for discussion 103. Answer: 1{3.
(c) Trick question! This limit does not exist.
(d) We can calculate directly.
lim
Ñ2
x
p x4
x2 q p16
x2
4q
x4 x2 20
xlim
Ñ2
x2
2
px 5qpx2 4q
xlim
Ñ2
x2
xlim
p
x2 5qpx 2q
Ñ2
9 4 36.
To check your work, you can observe that f 1 pxq 4x3
2
2x and then calculate f 1 p2q 4 8
4 36.
(e) Compare to problem 2 on quiz 3 for discussion 102. Answer 1{4.
(f) Compare to problem 2 on quiz 2 for discussion 103. Answer 1{16.
Problem 5.
35 cospxq 950 has a solution.
Show that the equation cospxq x4 has a solution in the interval p0, 1q.
Let f pxq ex 5 sinpxq. Show that there exists some c such that f pcq 50.
(a) Show that the equation x2
(b)
(c)
Solution.
(a) Observe that x2 35 cospxq 950 if and only if x2 35 cospxq 950. So we are trying to show that, if
g pxq x2 35 cospxq, there is a c such that g pcq 950. Observe that g is continuous, g p0q 0 35 950,
and g p950q 9502 35 cosp950q ¡ 950. Thus, by the intermediate value theorem, g pcq 950.
(b) cospxq x4 has a solution in p0, 1q if and only if hpxq x4 cospxq has a root in p0, 1q. Observe that
hpxq is continuous, hp0q 1 0, and that hp1q 1 cosp1q ¡ 0 (why?). Thus by the intermediate
value theorem there exists a c such that hpcq 0.
(c) Observe that f pxq is continuous, f p0q 1 50, and f p50q
intermediate value theorem, there is a c such that f pcq 50.
e50
5 sinp50q
¡
50. Thus by the
Problem 6. Find a number δ ¡ 0 such that the stated condition is satisfied.
(a) x2 1{9 whenever |x| δ.
(b) |1{x 1{3| 1 whenever |x 3| δ.
Solution.
(a) Observe that x2 x2 1{9 implies that |x|
|x| δ 1{3, we do get x2 1{9.
1{3, suggesting we try δ 1{3, and indeed, for any
(b) Observe that
1
x
13 1 ðñ 1 x1 13 1 ðñ 32 x1 43 .
Since we trying to find an interval around 3, we can ignore negative values of 1{x (draw a picture!).
Thus, we have 1{x 4{3, which implies that x ¡ 4{3. Working backwards, we can show that whenever
x ¡ 4{3, we get |1{x 1{3| 1. Thus, if we let δ 3 4{3 5{3 (or anything smaller), and then for
every x such that |x 3| δ 4{3, we get |1{x 1{3| 1, as desired.
Problem 7. Prove the following using the precise definition of limits.
(a) lim p2x
x
Ñ3
(b) lim px
(c)
x
Ñ8
x
lim e
Ñ8
5q 1.
sinpxqq 8.
x
0.
Solution.
3
(a) Compare to problem 3 on quiz 2 for discussion 102.
(b) Let M ¡ 0 be given. Let N M
with this!). Then whenever x ¡ N
Thus, by definition, lim px
x
1 (make sure you understand how one would go about coming up
M 1, we have
x sinpxq ¡ N 1 pM
sinpxqq 8.
Ñ8
1q 1 M.
(c) Let ¡ 0 be given, and let N lnpq. Then whenever x
|ex | |ex 0| . Thus, by definition, lim ex 0.
N , we have ex eN elnpq , so
Ñ8
x
Problem 8. Use the definition of the derivative; do not use any differentiation rules.
(a) Let f pxq x3 2
. Compute f 1 p2q.
x 1
(b) Let g pxq x2
2. Find an equation for the tangent line at x 1.
Solution.
(a) Compare to problem 3 on quiz 3 for discussion 102.
(b) The slope of the line is given by
g 1 p1q lim
x
Ñ1
p x2
2q 3
x1
xlim
x2 1x 1 lim px
Ñ1
xÑ1
Thus the equation of the line is y 3 2px 1q, which is y
2x
1q 2.
1.
Problem 9. Feel free to use differentiation rules.
d
(a) Compute
dx
?x
x
1
.
(b) Find equations for the tangent line to the curve y
(c) Find equations for the tangent lines to the curve y
ex3x 1 at x 1.
3x3 6x which are parallel to the line y 18x
Solution.
(a) Note that you do not yet have
? a rule for computing the derivative of
calculate the derivative of x.
2.
?x. So we first must use limits to
?x ?a
?x ?a
1
1
?
?
?
?
?
lim ?
lim ?
.
xÑa
xÑa p x xa
aqp x
aq xÑa x
a
2 a
?
?
Thus, the derivative of x is 1{p2 xq. We can now use the quotient rule to compute
derivative:
?
1
? px 1q ? x
d
x
2 x
2x?xp1x 2x
2?xpxx 11q2 .
dx x 1
px 1q2
1q2
lim
4
the desired
(b) The derivative is given by
x
x
pe pe1xq31q3x2 pe q
and at x 1, this evaluates to pp3e 3q 3eq{pe 1q2 3{pe 1q2 .
y1
Now use point-slope form or
whatever other method you wish to calculate the equation of the tangent.
2
(c) The slope of the curve is given by y 1 9x2 6.
a This is parallel to the line 18x 2 whenever 9x 6 18,
or 9x2 24, or x2 24{9 8{3, or x 8{3 (sorry, I meant for these numbers to turn out nicer,
but I made an arithmetic error while coming up with the problem it seems—still, its not unbearably
messy, I think). You know the slope at both of these points is 18, so now, you just have to compute the
y-intercepts for the tangents at both these points.
Problem 10.
(a) Sketch the graph of any function f pxq such that lim f pxq
x
Ñ8 5.
and lim
x
Ñ8
8, xlim
f pxq 8, lim f pxq 8
Ñ0
xÑ0
(b) Sketch the graph of any differentiable function g pxq such that lim g 1 pxq 1 and lim g 1 pxq 1.
(c) Sketch the graph of the function hpxq Ñ8
x
a
tanpxq.
Solution. I don’t know how to include graphs here.
5
x
Ñ8