Quiz, Math 1A
Friday, September 9, 2011
Name:
Student ID#:
Put personal items under your seat. NO USE OF NOTES, TEXTS,
CALCULATORS, OR FELLOW STUDENTS IS ALLOWED
Each problem is worth 3 points.
Show all of your work in order to receive full credit.
1. Let f (x) =
√
2x + 1.
(a) Explain why f is one-to-one.
(b) Find f −1 and give its domain.
2. Simplify the following expression:
log6 (4) + log3 (9log6 (3) ).
3. Evaluate the difference quotient
f (a + h) − f (a)
h
for f (x) = x2 , h 6= 0.
1
Solutions:
√
1. (a) f (x) = 2x + 1 is just a compression of the square root function
by a factor of two followed by a left shift of one. These operations
preserve one-to-one-ness, so since the square root function is oneto-one, we see that f is also one-to-one.
(b) To find f −1 , we calculate:
x = f (y)
p
x = 2y + 1
x2 = 2y + 1
x2 − 1 = 2y
(x2 − 1)/2 = y
This gives us the inverse function f −1 (x) = (x2 − 1)/2. The
range of f is the interval [0, ∞) since 2x + 1 has a range of all
real numbers, and so f achieves the full range of the square root
function. The domain of f −1 is just the range of f , which is
[0, ∞).
2. We calculate:
log6 (4) + log3 (9log6 (3) ) = log6 (4) + log6 (3) log3 (9) = log6 (4) + log6 (3) log3 (32 )
= log6 (4) + log6 (3) ∗ 2 log3 (3) = log6 (4) + 2 log6 (3)
= log6 (4) + log6 (32 ) = log6 (36) = log6 (62 ) = 2 log6 (6) = 2
3. We evaluate:
f (a + h) − f (a)
(a + h)2 − a2
a2 + 2ah + h2 − a2
=
=
h
h
h
2
2ah + h
(2a + h)h
=
=
= 2a + h
h
h
2