MATH 1A, FALL 2012
HOMEWORK 1 SOLUTIONS
WRITTEN BY SHISHIR AGRAWAL
Sec 1.1, ex 36. Find the domain of the function
g(t) =
√
3−t−
√
2 + t.
We require that the expressions under the radical be nonnegative. Thus, we want 3 − t ≥ 0,
which is equivalent to t ≤ 3, and we want that 2 + t ≥ 0, which is equivalent to t ≥ −2. In
other words, the domain is the closed interval [−2, 3].
Sec. 1.1, ex. 76. Determine whether f (x) = x |x| is even, odd, or neither.
Observe that f (−x) = (−x) |−x| = −x |x| = −f (x), so this function is odd.
Sec. 1.2, ex 8. Determine expressions for the quadratic functions f and g described. (a) The graph of f is
an upwards-facing parabola with has its vertex at (3,0) and passes through the point (4, 2).
(b) The graph of g passes through the points (-2, 2), (0, 1) and (1, -2.5).
For (a), we can use a geometric approach. Consider first the graph of the function f1 (x) = x2 .
We first want to translate the graph of this function to the right 3 units, and if we call this
translate f2 , its equation is
f2 (x) = f1 (x − 3) = (x − 3)2 .
Now observe that f2 (4) = 1, but since (4, 2) is on the graph of f , we need f (4) = 2. So, to
get from f2 to f , we vertically scale the graph by a factor of 2. In other words,
f (x) = 2f2 (x) = 2(x − 3)2 .
For (b), we proceed more algebraically. The function f must be given by an equation of the
form
f (x) = ax2 + bx + c.
Now we know that (−2, 2) is on the graph of f , so we must have f (−2) = 2, which means
that
2 = a(−2)2 + b(−2) + c = 4a + 2b + c.
Similarly, plugging in the points (0, 1) and (1, −2.5), we get two additional equations.
1=c
−2.5 = a + b + c
Now we solve this system of three equations in the three unknowns a, b, and c, and we find
that c = 1, b = −2.5 and a = −1. Thus,
f (x) = −x2 − 2.5x + 1.
1
2
WRITTEN BY SHISHIR AGRAWAL
Sec 1.2, ex 12. The manager of a weekend flea market knows from past experience that if he charges x
dollars for a rental space at the market, then the number of spaces y that he can rent is
given by the equation
y = 200 − 4x.
(a) Sketch the graph of this linear function. (Remember that the rental charge per space
and the number of spaces rented cannot be negative quantities.) (b) What do the slope, the
y-intercept, and the x-intercept of the graph represent?
For part (a), the graph is linear with y-intercept 200, slope −4 and x-intercept 50. We
restrict this line to only the first quadrant.
For part (b), the slope is the change in the number of spaces he can rent for every dollar
he increases the price of a rental space. The y-intercept is the number of spaces he can rent
if he gives out rental spaces for free. The x-intercept is the maximum price he can charge
before no one is willing to rent spaces from him anymore.
Sec 1.2, ex 26. The number of species S living in caves in central Mexico has been related to the surface
area A of the caves by the equation S = 0.7A0.3 . (a) If a cave has surface area A = 60 m2 ,
how many species would you expect to find in that cave? (b) If you discover that 4 species
of bats live in a cave, estimate the area of the cave.
For part (a), we simply calculate S = 0.7(60)0.3 = 2.4. Thus, we expect about 2 species of
bats in this cave.
For part (b), we know that S = 4 and we want to find the corresponding value of A by
solving the equation 4 = 0.7A0.3 .
4 = 0.7A0.3
7
4=
A3/10
10
40
= A3/10
7
40
7
10/3
= (A3/10 )10/3 = A
Thus, the area A of the cave is approximately 333.6 m2 .