Download Quiz 2 - Purdue Math

Quiz 2
Histograms
1
Solution
Problem 1. Given f (x) = x − 1 and g(x) = x2 − 1, find all the numbers x such that
f (g(x)) = g(f (x)).
Solution. To set up the equation, let us find f (g(x)) and g(f (x)) first. We have
f (g(x)) = g(x) − 1 = (x2 − 1) − 1 = x2 − 2
and
g(f (x)) = [f (x)]2 − 1 = (x − 1)2 − 1 = (x2 − 2x + 1) − 1 = x2 − 2x.
Equating them,
f (g(x))
x2 − 2
−2
x
=
=
=
=
g(f (x)), i.e.,
x2 − 2x, i.e.,
−2x, namely
1
So x = 1 is the only number satisfying f (g(x)) = g(f (x)).
Problem 2. Find the difference quotient DQ =
−2x2 + 3x.
Solution. For the given f (x),
DQ =
=
=
=
=
=
=
f (x + h) − f (x)
of the function f (x) =
h
f (x + h) − f (x)
h
[−2(x + h)2 + 3(x + h)] − (−2x2 + 3x)
h
[−2(x2 + 2xh + h2 ) + 3(x + h)] + 2x2 − 3x
h
2
+ 3h + −2x
− 4xh − 2h2 + 3x
2x2 + −3x
h
−4xh − 2h2 + 3h
h
h(−4x − 2h + 3)
h
−4x − 2h + 3
2