Math 165: Calculus 1
Problem Set 1: Fall 2024
Due Date: Tuesday, September 3 by 11:59pm
Printed Name
Signed Name
Instructor Name
TA Name
Directions. Show all work and justify all answers for full credit. Your solutions will be graded for
neatness, clarity, conciseness, and correctness. Only hand-written solutions will be accepted. Upload a
PDF scan of your solutions by the speciÖed due date.
1
1. Consider the function f (x) = ex shown below.
y
10
9
8
7
6
5
4
3
2
1
-3
-2
-1
0
1
2
3
x
(a) On the graph shown above, sketch the four lines whose slopes are given by
(i) fARC [0; 2], (ii) fARC [0; 1], (iii) fARC [0; 0:5], (iv) fIRC (0):
(b) Use numerical methods to estimate the instantaneous rate of change fIRC (0) of the function
f (x) = ex at the point where x = 0: You can do this by completing the table below. Round
your answers to Öve decimal places.
h>0
fARC [0; h]
0:5
0:1
0:01
0:001
According to the table, it appears that the slope of the tangent line to the curve f (x) = ex at
the point where x = 0 is given by
fIRC (0) = lim fARC [0; h] =
h!0+
2
:
(c) Fix any real number a and let h > 0: Use properties of the exponential and basic algebra to
show that if f (x) = ex , then
fARC [a; a + h] = ea fARC [0; h]:
Hint: Work out the right and left-hand sides separately until you get the same expression for
both.
(d) Use the results from (b) and (c) above to Önd a formula for the instantaneous rate of change (i.e.,
the slope of the tangent line) of f (x) = ex at the point where x = a: That is, what happens
to fARC [a; a + h] when h is very close to 0? Your answer will be a very simple expression that
depends on the real number a:
fIRC (a) = lim fARC [a; a + h] =
:
h!0+
(e) Use the formula from the previous exercise to Önd the slope of each tangent line to the curve
f (x) = ex at each requested point where x = 0; 1; 2. Give the exact answer (some power of e
will do).
fIRC (0) =
fIRC (1) =
3
fIRC (2) =
2. Consider the function f (x) = x4 and let a be any real number
(a) Find a simpliÖed expression for the di§erence quotient fARC [a; a + h]: Your answer should be an
expression in terms of a and h. Be sure to cancel the troublesome h in the denominator.
fARC [a; a + h] =
4
(b) Find a very simple formula (that depends on the real number a) for the slope fIRC (a) of the
tangent line to the curve at f (x) = x4 the point where x = a: In other words, use the expression
from part (a) to Önd fARC [a; a + h] when h is very close to 0:
fIRC (a) = lim fARC [a; a + h] =
:
h!0+
(c) Use the formula from the previous exercise to Önd the slope of the tangent line to the curve
f (x) = x4 at each point where x = 0; 1; 2.
fIRC (0) =
fIRC (1) =
5
fIRC (2) =
3. Consider the function f (x) = x5 and let a be any real number
(a) Find a simpliÖed expression for the di§erence quotient fARC [a; a + h]: Your answer should be an
expression in terms of a and h. Be sure to cancel the troublesome h in the denominator.
fARC [a; a + h] =
6
(b) Find a very simple formula (that depends on the real number a) for the slope fIRC (a) of the
tangent line to the curve at f (x) = x5 the point where x = a: In other words, use the expression
from part (a) to Önd fARC [a; a + h] when h is very close to 0:
fIRC (a) = lim fARC [a; a + h] =
:
h!0+
(c) Compare the result in part (b) with the results in Exercise 2(b) above and Exercise 2(b) of the
Sample Problem Set 1 Solutions. You have found a formula for the slope fIRC (a) of the tangent
line to the curve at f (x) = xn the point where x = a for n = 3; 4; 5: Guess a formula for the
slope fIRC (a) of the tangent line to the curve at f (x) = xn the point where x = a where n is
any positive integer. You do not need to show any work. This is where the Power Rule for
Derivatives comes from!
fIRC (a) = lim fARC [a; a + h] =
h!0+
7
:
4. Assume that h > 0 and compute a simpliÖed expression for the average rate of change for each function
f on the given interval [a; a + h]. Be sure to cancel the troublesome factor of h from the denominator
in each case. This way, you can easily use the expression fARC [a; a + h] to guess the slope fIRC (a) of
the tangent line at the point where x = a:
(a) If f (x) = x1 and a = 2, then
fARC [2; 2 + h] =
and fIRC (2) =
8
(b) If f (x) =
p
x and a = 4, then
fARC [4; 4 + h] =
and fIRC (4) =
Hint: Use a conjugation trick from high school to simplify.
9
:
5. Let
f (x) =
cos(x)  1
:
x
Use numerical methods to compute
lim f (x) =
:
x!0
You can do this by completing the table below.
x>0
f (x)
x<0
0:5
0:5
0:1
0:1
0:01
0:01
0:001
0:001
10
f (x)
6. Use the previous exercise and the methods outlined in the "Sample Problem Set 1 (Solutions)" to
Önd a formula for the instantaneous rate of change fIRC (a) of the function f (x) = cos(x) at any point
where x = a: Show all work in the space below.
fIRC (a) =
11
7. On the set of coordinate axes below, sketch the graph of the following piece-wise deÖned function.
Indicate the gluing points with open circles and closed dots as appropriate. Use your sketch to Önd
each quantity. If the quantity does not exist, write "DNE" in the answer box.
8
x
x<1
>
>
<
3
x=1
f (x) =
2
2

x
1
<
x<2
>
>
:
x3
x2
y
4
3
2
1
-4
-3
-2
-1
1
2
3
4
x
-1
-2
-3
-4
(a)
lim f (x) =
x!1
(b)
lim f (x) =
x!1+
(c) lim f (x) =
(d) f (1) =
x!1
(e) lim f (x) =
(f) lim f (x) =
(g) lim f (x) =
(h) f (1) =
(i) lim f (x) =
(j) lim f (x) =
(k) lim f (x) =
(l) f (2) =
x!1
x!2
x!1
x!1+
x!2
x!2+
12
8. Write the function
jx  1j
x2  1
as a simpler piece-wise deÖned function and graph it on the set of coordinate axes provided.
f (x) =
y
4
3
2
1
-4
-3
-2
-1
1
2
3
4
x
-1
-2
-3
-4
Use your graph compute each quantity graphically.
lim f (x) =
x!1
(b) lim f (x) =
(c) lim f (x) =
x!1
x!1+
13
(d) f (1) =
9. Write the function
f (x) =
x2  jx  1j  1
:
jx  1j
as a simpler piece-wise deÖned function and graph it on the set of coordinate axes provided.
y
4
3
2
1
-4
-3
-2
-1
1
2
3
4
x
-1
-2
-3
-4
Use your graph compute each quantity graphically.
lim f (x) =
x!1
(b) lim f (x) =
(c) lim f (x) =
x!1
x!1+
14
(d) f (1) =
10. Sketch the graph of f (x) = x  bxc where bxc denotes the greatest integer function (or áoor function)
and Önd each of the following quantities if they exist. Note: You may have also seen the notation
[[x]] to denote the same function),
y
4
3
2
1
-4
-3
-2
-1
1
2
3
4
x
-1
-2
-3
-4
Use your graph compute each quantity graphically.
(a) f (0) =
(e) lim f (x) =
x!0+
(b) f (1) =
(c) f (5:25) =
(f) lim f (x) =
(g) lim f (x) =
x!0
x!0
15
(d) f (7:25) =
(h)
lim f (x) =
x!5:25