MATHEMATICS 120 PROBLEM SET 6 – SOLUTIONS
1. A ball is thrown upward from the ground and falls back to the ground after 8 seconds.
What was its initial velocity? (The gravitational acceleration is 9.8 m/s2 .)
Let v(t) denote the velocity of the
R ball at time t (the positive direction is upward). Then
0
v (t) = g = −9.8, so that v(t) = −9.8dt = −9.8t + C. If the ball falls back to the ground
after 8 seconds, it reaches its highest point after 4 seconds, and at that point the velocity
is 0. Therefore 0 = v(4) = −9.8 · 4 + C, C = 39.2, so that v(t) = −9.8t + 39.2. The initial
velocity is v(0) = 39.2m/s.
2. Let f (x) be a function such that
and find (f −1 )0 (0).
(f (x))3
= x5 + 4x. Prove that f (x) is one-to-one,
(f (x))2 + 1
Suppose that f (x1 ) = f (x2 ), then we also have x51 + 4x1 = x52 + 4x2 . We have to prove
that this can happen only if x1 = x2 . To do this, it suffices to check that the function
g(x) = x5 + 4x is one-to-one. But g 0 (x) = 5x4 + 4 > 0 for all x, therefore g(x) is increasing,
therefore it is one-to-one.
x3
= y 5 + 4y. Differentiate this implicitly:
Let y = f −1 (x), then 2
x +1
3x2 (x2 + 1) − x3 · 2x
= 5y 4 y 0 + 4y 0 ,
(x2 + 1)2
x4 + 3x2
= (5y 4 + 4)y 0 .
2
2
(x + 1)
5
If x = 0 then y + 4y = 0, so y = 0. Plug in x = y = 0: 0 = 4y 0 , hence y 0 (0) = 0.
3. Prove that if f (x) and g(x) are one-to-one functions defined for all x, then the composite
function f (g(x)) is also one-to-one.
Suppose that f (g(x1)) = f (g(x2)). Since f is one-to-one, we must have g(x1 ) = g(x2 ).
Since g is one-to-one, it follows that x1 = x2 . Therefore f (g(x)) is one-to-one.
4. Solve the equation log9 (x + 1) − log81 (x + 7) =
1
.
4
We simplify the equation:
log(x + 1) log(x + 7)
1
−
= ,
log 9
log 81
4
log(x + 1) log(x + 7)
1
−
= ,
2 log 3
4 log 3
4
2 log(x + 1) − log(x + 7) = log 3,
(x + 1)2
log
= log 3.
x+7
(x + 1)2
= 3, x2 + 2x + 1 = 3x + 21, x2 − x − 20 = 0, x = −4 or x = 5. But if
x+7
x = −4, then log9 (x + 1) = log9 (−3) is not defined. Therefore the only solution is x = 5.
Hence
1