Sample Exam 1 Solutions
1a.
(1) This is a parabola shifted up one unit, so the vertex is at (0,1)
(2) This is a straight line. The x-intercept is found by setting 2x−3 = 0, so
x= 1.5. The y-intercept is found by setting x= 0, so y=−3
(3) This is the graph of √xshifted right by 10 units.
(4) This is the graph of |x|shifted right 3 and up 2
1b. Write k(x) = f(x) + g(x) where f(x) = √x−10 and g(x) = 3
x−10 . Then
dom(f) is xso that x−10 ≥0, which gives the interval [10,+∞). Also, dom(g) is
everything but 10; the intersection of the two will then be (10,∞)
1c. We have the average rate of change is
q(8) −q(5)
8−5=(82+ 2(8) −4) −(52+ 2(5) −4)
3
(1)
=76 −31
3
(2)
=45
3= 15(3)
2a. The key thing to remember here is that a point (x, y) is on the graph of
f(x) if and only if f(x) = y
(1) What the point with x-coordinate 1? The point (1,3). So we must have
f(1) = 3
(2) (f+g)(3) = f(3) + g(3) = 2 + 1 = 3.
(3) (f/g)(2) = f(2)/g(2) = 2.5/0: Undefined!
(4) (f◦g)(−3) = f(g(−3)) = f(−1) = 4
(5) (g◦g)(−3) = g(f(−3)) = g(4) = 1
2b. We need to test whether or not (c◦d)(x) = xand (d◦c)(x) = x. We get
(c◦d)(x) = c(x
2−5) = 2( x
2−5) + 5 = (x−10) + 5 = x−5. Since this isn’t x, the
functions are not inverses.
Note that if you’re a little careless here you might right c(d(x)) = 2 x
2−5 + 5 = x
and mistakenly conclude that the functions are inverses. The moral: always use
parentheses!
2c.
(1) Replace f(x) with f(x) + 5 = 3
x+ 15. Let’s call this g(x)
(2) Replace g(x) with 2g(x) = 6
x+ 30. Call it h(x)
(3) Replace h(x) with h(x−7) = 6
x−7+ 30.
3a. Let’s rewrite this as:
10x+ 30 ≤5 When 10x+ 30 ≤0(4)
−(10x+ 30) ≤5 When 10x+ 30 >0(5)
(6)
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