Download Math 119 – Midterm Exam #1 (Solutions)

Math 119 – Midterm Exam #1 (Solutions)
1. Determine the domain of the functions below.
√
40 − x
.
(a) f (x) =
x−3
We must have x 6= 3 to avoid division by 0. We also must avoid
taking the square root of a negative number and for that we need
x < 40. The domain of f is therefore the set of all real numbers
x which satisfying x 6= 3, x < 40.
r
p
1
(b) g(y) = y + 5 −
.
y−1
To avoid negative numbers in the first square root we must have
y ≥ −5. To avoid negative numbers and division by 0 in the
second square root we need y > 1. Therefore the domain of g is
the set of all reals y satisfying y > 1.
2. Simplify the difference quotient
f (x + h) − f (x)
for f (x) = x3 .
h
Simple arithmetic yields
3x2 + 3xh + h2 .
3. Graph the function f (x) = x2 − 4x − 5, find the y-intercept, the xintercepts and determine the minimum value that f attains.
The y intercept is (0, f (0)) = (0, −5). The x-intercepts may be obtained by factoring the polynomial since f (x) = (x − 5)(x + 1). The
x-intercepts are (−1, 0) and (5, 0).
The minimum value of f is attained at the vertex of the parabola,
namely, at (2, −9). Hence the minimum value is −9.
4. Compute the limits below (justify steps).
x2 + 9
x→2 x3 + 3x
This is a rational function and 2 is in its domain. Since rational functions are continuous, it is enough to substitute x by 2,
yielding 13/14 as the limit.
(a) lim
x2 + 4x + 2
x2 (1 + 4/x + 2/x2 )
1 + 4/x + 2/x2
=
=
lim
=
lim
x→+∞ 3x2 + 9x + 100
x→+∞ x2 (3 + 9/x + 100/x2 )
x→+∞ 3 + 9/x + 100/x2
1
3
q
√
(c) lim x + x
(b)
lim
x→1
We will use the power (limit) rule twice. Notice that
√
√
lim (x + x) = lim x + lim x = 2.
x→1
x→1
Hence
lim
x→1
x→1
q
q
√
√
√
x + x = lim (x + x) = 2.
x→1
5. Consider the function
(
x2 + 137432954x − 2,
f (x) =
4x − 2,
if x > 0
if x ≤ 0.
Determine the values of x for which f is continuous (justify). Is there
a value of x between −10 and 1 for which f (x) = 2010?
The function f is linear for x ≤ 0 and a polynomial for x > 0. Therefore we know that it is continuous everywhere, except perhaps at x = 0.
To check whether f is continuous at x = 0 we have to compute the
side limits
lim f (x) = lim (4x − 2) = −2
x→0−
x→0−
and
lim f (x) = lim (x2 + 137432954x − 2) = −2.
x→0+
x→0+
Therefore
lim f (x) = −2 = f (0)
x→0
and thus f is continuous at x = 0.
Notice that f is continuous on the interval [−10, 1] and
f (−10) = 4 · (−10) − 2 = −42 < 2010
2
f (1) = (1)2 + 137432954(1) − 2 = 137432953 > 2010.
By the Intermediate Value Property, there exists some x, −10 < x < 1
such that f (x) = 2010.
√
6. Find the derivative of f (x) = x. Determine the equation of the line
tangent to the graph of f√ when x = 4. Use the linear equation to
approximate the value of 4.01.
By the power rule, f 0 (x) =
1
√
.
2 x
The slope of the tangent line at x = 4 is f 0 (4) = 41 . The tangent
line must contain the point (4, f (4)) = (4, 2) and therefore the line
equation is given by L(x) = 14 x + 1.
√
The value L(4.01) = 2.0025 is a good approximation for 4.01 =
2.00249843945 . . . .
9. A company uses a truck to deliver its products. To estimate the costs
the manager models fuel consumption by the function
G(x) =
1 1200
+x
250
x
which gives the use of diesel in gal/mile assuming that the truck is
driven at constant speed of x miles per hour (x ≥ 5).
The driver is paid $20/hour to drive 250 miles and diesel costs $4/gallon.
• Find the cost function C(x). (Simplify the expression as much as
possible.)
The total distance is 250 miles. At a speed of x mph, the total
amount of fuel used is 250 · G(x) gallons. At $4 per galon, the
total fuel cost is F (x) = 4 · 250 · G(x). Simplifying this yields
F (x) = 4
1200
x
+x .
The driver is paid by the hour and the number of hours he must
drive is 250/x. Hence the total cost (fuel plus driver) is given by
C(x) =
1200
9800
20 · 250
+4
+x =
+ 4x.
x
x
x
3
• At what rate is the cost C(x) changing with respect to x when the
truck is driven at 40mph? Is the cost increasing or decreasing at
that speed?
The question amounts to calculating C 0 (40). Since C 0 (x) = 4 −
9800
9800x−2 we get C 0 (40) = 4 − 1600
< 0. Because the derivative
is negative we can tell that the cost is decreasing at that speed
(increasing the speed a little bit beyond 40mph will cause the
cost to decrease).
4