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Math 105: Review for Exam I - Solutions
- Let f(x) = 3 +
x + 5
f(x) = 3 +
x + 5 f(x) = 3 +
x + 5.
(a) What is the natural domain of fff? [− 5 , ∞), which means all reals greater than or equal to 5
(b) What is the range of fff? [3, ∞), which means all reals greater than or equal to 3
- For the graph of fff shown, answer the following.
(a) Evaluate the following.
i. f
′
f(−2)
′
f(−2)
′
ii. fff(3)(3)(3) = 2
iii. lim
x→ 3
−
lim f(x)
x→ 3
−
lim f(x)
x→ 3
−
f(x) = − 2
iv. lim
x→ 3
f(x)
lim
x→ 3
f(x) lim
x→ 3
f(x) = 3
v. lim
x→ 3
limf(x)
x→ 3
limf(x)
x→ 3
f(x) does not exist
vi. lim
x→ 2
limf(x)
x→ 2
limf(x)
x→ 2
f(x) = 0
(b) Where is fff discontinuous? at x = 3
(c) Where does f
′
f
′
f
′
fail to exist?
at x = − 3 , − 1 , 0 , 3
2
4
−
−4 −2 2 4
f(x)
3
3
- Let f(x) = 3x
2
− 2 x
f(x) = 3x
2
− 2 x f(x) = 3x
2
− 2 x.
(a) Compute the average rate of change of fff on the interval [2,2.1].
f(2.1) − f(2)
(b) Using the limit definition of the derivative, find f
′
f(x)
′
f(x)
′
(x).
f
′
(x) = lim
h→ 0
f(x + h) − f(x)
h
provided this limit exists
= lim
h→ 0
3(x + h)
2
− 2(x + h) − (3x
2
− 2 x)
h
= lim
h→ 0
3 x
2
2
− 2 x − 2 h − 3 x
2
h
= lim
h→ 0
h(6x + 3h − 2)
h
= lim
h→ 0
(6x + 3h − 2)
= 6x − 2
(c) Find the equation of the tangent line to fff at xxx = 2= 2= 2.
We want y = mx + b. m = f
′
(2) = 6 · 2 − 2 = 10, so y = 10x + b.
When x = 2, y = f(2) = 3 · 2
2
Thus, 8 = 10 · 2 + b, so b = −12 and we have y = 10x − 12.
(d) How would the derivative of ggg(((xxx) =) =) = fff(((xxx) + 5) + 5) + 5 compare to f
′
f(x)
′
f(x)
′
(x)?
The graph of y = f(x) + 5 is the graph of y = f(x) shifted vertically by 5 units, but this has no
effect on the slope of the graph, so g
′
(x) = f
′
(x).
(e) How would the derivative of h(x) = 5f(x)
h(x) = 5f(x) h(x) = 5f(x) compare to f
′
(x)
f
′
(x) f
′
(x)?
The graph of y = 5f(x) is the graph of y = f(x) stretched vertically by a factor of 5; this also
results in slopes that are 5 times greater at any given x-value. Thus, h
′
(x) = 5f
′
(x).
Note that we get the same result by considering our derviative rule
d
dx
[kf(x)] = kf
′
(x) where
k = 5.
- Fill in the table showing the graphical relationships between fff , f
′
f
′
f
′
, and f
′′
f
′′
f
′′
fff positive negative increasing decreasing concave up concave down
f
′
f
′
f
′
X X positive negative increasing decreasing
f
′′
f
′′
f
′′
X X X X positive negative
- Given the graph of f
f f , sketch a graph of f
′
f
′
f
′
and a graph of F
F
F , an antiderivative of f
f f such
that F (0) = − 2
F (0) = − 2
F (0) = − 2.
F(x)
f(x)
o
o
o
o
f ‘(x)
Note: The concave up portion in the middle of the graph of f is a perfect parabola, so its derivative
(f
′
) is linear; since you don’t know the equation for f, your graph of f
′
may be concave up/down there.
(b) Based on the data, is f
′′
f
′′
f
′′
(1.0) positive or negative?
Using the same procedure as in the previous part, we can make the following estimates.
f
′
(0.85) ≈ 12 f
′
(0.95) ≈ 11 f
′
(1.05) ≈ 9 f
′
We see from these estimates that f
′
(x) appears to be decreasing near x = 1. If f
′
(x) is decreasing,
then f
′′
(x) is negative (that is, f(x) is concave down).
- Find the derivatives of the following.
(a) y = 2 + 3x + x
4
6
y = 2 + 3x + x
4
6
y = 2 + 3x + x
4
6
y
′
= 3 + 4x
3
5
(b) y =
6
x +
x
6
x
x
π
1 / 2
y =
6
x +
x
6
x
x
π
1 / 2
y =
6
x +
x
6
x
x
π
1 / 2
First, rewrite y to make it easier to apply our derivative rules:
y = x
1 / 6
− 6
· x + 6x
− 1
π
1 / 2
y
′
x
− 5 / 6
− 7
− 2
If necessary, we can rewrite this using exponent rules: x
−a
= 1/x
a
and x
a/b
b
a.
y
′
6
x
5
x
7
x
2
- Find antiderivatives of the following.
(a) y = π + 3x
2
y = π + 3x
2
y = π + 3x
2
antiderivative = πx + x
3
+ C
(b) y = 4x
5
x
6
y = 4x
5
x
6
y = 4x
5
x
6
= 4x
5
− x
− 6
antiderivative =
4 x
6
x
− 5
+ C =
2 x
6
5 x
5
+ C
