1
MATH 114 Test 1 Solutions
1. For the function
 2
if x ≤ 0;
 x + 2,
2 − x,
if 0 < x ≤ 2;
f (x) =

(x − 2)2 , if x > 2.
a) Draw the graphs of the functions y = f (x) and y = 2f (x − 2).
Solution:
4
4
3
3
f
2
1
-4
-3
-2
-1
0
0
2f(x-2)
2
1
1
2
3
0
0
4
-1
-1
-2
-2
-3
-3
-4
-4
1
2
3
4
5
b) State the domain and range of y = f (x).
Solution: Domain = all real numbers. Range = all nonnegative numbers.
c) Is f an invertible function? Explain.
Solution: No. The horizontal line y = 1 cuts the graph in two points. Therefore
f cannot be invertible by the horizontal line test.
2. Solve each equation for x.
a) ln(x) + ln(x + 5) = − ln(1).
Solution: ln(x(x + 5)) = − ln(1) = 0. Therefore,
x(x + 5) = 1, and x2 + 5x − 1 = 0.
√
Using the quadratic formula, x = −5±2 29 .
x
b) ee = 4.
Solution: Taking the natural log of both sides of the equation, we get ex = ln(4).
Taking the natural log once more, we get x = ln(ln(4)).
√
3. For f (x) = x2 , and g(x) = x − 1.
a) Find f ◦ g and g ◦ f.
√
√
Solution: f ◦ g = ( x − 1)2 = x − 1. g ◦ f = x2 − 1.
b) Find the domain and range of g ◦ f.
(4pts)
2
Solution: g ◦ f is defined only when x − 1 ≥ 0. This means that the domain is
|x| ≥ 1; that is, x ≤ −1 or x ≥ 1.
√
Given x2 − 1 can be any nonnegative number, it follows that y = x2 − 1 can be
any nonnegative number. Thus the range is all nonnegative values.
6
7
2
MATH 114 Test 1 Solutions
4. Let the graphs of f and g be given as below. Evaluate the limits:
4
4
f
3
-4
-3
-2
2
2
1
1
0
0
-1
3
1
2
3
4
-4
-3
-2
0
0
-1
-1
-1
-2
-2
-3
-3
-4
-4
g
1
2
3
a) lim+ f (x) + g(x)
x→2
Solution:
lim f (x) + g(x) = lim+ f (x) + lim+ g(x) = 2 + 1 = 3.
x→2+
x→2
x→2
b) lim− f (x)g(x)
x→2
Solution:
lim f (x)g(x) = lim− f (x) lim− g(x) = 0 · 1.8 = 0.
x→2−
x→2
x→2
c) lim f (x)g(x)
x→0
Solution:
lim f (x)g(x) = lim+ f (x) · lim+ g(x) = 2 · 0 = 0.
x→0+
x→0
x→0
lim f (x)g(x) = lim− f (x) · lim− g(x) = 3 · 0 = 0.
x→0−
x→0
x→0
Since the left and right limits both equal 0, it follows that lim f (x)g(x) = 0.
x→0
d) lim f (x)g(x)
x→2
Solution: The limit does not exists since lim− f (x)g(x) = 0 and
x→2
lim f (x)g(x) = lim+ f (x) · lim+ g(x) = 2 · 1.8 = 3.6
x→2+
x→2
x→2
The left and right limits are different.
e) At which points is h(x) = f (x)g(x) discontinuous? Explain.
Solution: Note that
lim h(x) = lim f (x)g(x) = 0 = h(0).
x→0
x→0
So h(x) is continuous at x = 0. However, limx→2 h(x) = limx→2 f (x)g(x) does not
exist, and hence h(x) is not continuous at x = 2. Also h(x) is continuous at all
other points, since both f and g are continuous at all other points.
4
3
MATH 114 Test 1 Solutions
5. A ball is thrown into the air with a velocity of 20 m/s, its height (in meters) after t
seconds is given by y = 20t − 5t2 .
a) Give a reasonable estimate of the instantaneous velocity of the ball at t = 2 using
average velocity.
Solution: We can use the average velocity between t = 2 and t = 2.1 seconds:
vav =
y(2.1) − y(2)
19.95 − 20
=
= −0.5m/s.
0.1
0.1
b) Find the instantaneous velocity of the ball at t = 2 seconds.
Solution: To find the instantaneous velocity, we evaluate the limit:
20t − 5t2 − 20
y(t) − y(2)
= lim
t→2
t→2
t−2
t−2
5(4t − t2 − 4)
−5(t − 2)2
= lim
= lim
t→2
t→2
t−2
t−2
= −5(t − 2) = 0.
lim
Therefore the instantaneous velocity is 0.
6. Suppose the graph of f is given below.
4
3
f
2
1
-4
-3
-2
-1
0
0
1
2
3
4
-1
-2
-3
-4
a) Sketch the graph of f ′ .
Solution: See the graph above.
b) For which points (if any), is f not differentiable? Explain.
Solution: f is not differentiable at x = 0 and x = 2 since the graph of f has
”sharp points” for these x-values.
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MATH 114 Test 1 Solutions
BONUS: Find points P and Q on the parabola y = 1 − x2 so the the triangle ABC formed
by the x-axis and the tangent lines at P and Q is an equilateral triangle (all sides equal).
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
–1 –0.8 –0.6 –0.4 –0.2 0
0.2 0.4 0.6 0.8
x
1
Solution: The angle CBA equals 60◦ or π3 radians. Therefore the slope of BC = − tan π3 =
√
√
2
′
′
− 3. The
derivative
of
the
function
y
=
1
−
x
is
y
=
−2x.
When
y
=
−
3, we get
√
√
√
√
−2x = − 3 and thus x = 23 . Thus P = ( 23 , 41 ). Similarly, Q = (− 23 , 14 ).