MTH131 Applied Calculus – Spring 2016
Lab 1 – SOLUTIONS
1
13
y = x−
3
3
(b) There are many solutions to each problem. The simplest comes when we set x = 0 and
get the corresponding y value. The table below gives that answer along with several
others for various values of x;
a)
b)
c)
(x1 , y1 ) equation
(x1 , y1 ) equation
(x1 , y1 ) equation
(0, 19)
y − 19 = 5x
(0, −2) y + 2 = 13 x
(0, − 32 ) y + 23 = − 35 x
1. (a)
a)
y = −3x + 6
b)
y = 5x + 19
c)
(1, 24)
y − 24 = 3(x − 1)
(3, −1)
y + 1 = 13 (x − 3)
(5, − 11
3 )
y+
11
3
(−3, 4)
y − 4 = 5(x + 3)
(6, 0)
y = 13 (x − 6)
(−5, 37 )
y−
7
3
2. (a) 3x1/2
(b) 3x−1/2
(c) 3x−2
(d) 9x−1
(e) 18x3/2
(f) 2x−2
3.
a)
31/2
b)
30
c)
3−1/2
3−1
d)
4. (a) 27y 4
(b) u2 v 2 w2 or (uvw)2
5. (a) The two compositions are determined as follows:
f (g(x))
=
=
=
f (x + 1)
(x + 1) + 1
(x + 1) − 1
x+2
2
and
x+1
g
x−1
x+1
+1
x−1
x+1 x−1
+
x−1 x−1
x + 1 + (x − 1)
x−1
2x
.
x−1
g(f (x))
=
=
=
=
=
= 3(x − 5)
= 5(x + 5)
(b) The most obvious set of functions is
f (x) = 2x2 + 17 and g(x) =
√
x
Using these, we have
√
f (g(x)) = f ( x)
√
= 2( x)2 + 17
= 2x + 17.
(c) There are many different answers here (an infinite number in fact). One possibility is
√
f (x) = 2x2 and g(x) = x + 17
Using these, we have
√
f (g(x)) = f ( x + 17)
√
= 2( x + 17)2
= 2(x + 17)
= 2x + 34.
6. The first 5 can all be solved by factoring:
(a) x2 − 7x + 12 = 0:
x2 − 7x + 12 = 0
(x − 3)(x − 4) = 0 ⇒ x = 3, 4
(b) x2 + 7x + 12 = 0:
x2 + 7x + 12 = 0
(x + 3)(x + 4) = 0 ⇒ x = −3, −4
(c) x2 − 7x + 12 = 2:
x2 − 7x + 12 = 2
x2 − 7x + 10 = 0
(x − 2)(x − 5) = 0 ⇒ x = 2, 5
(d) x2 + 3x − 18 = 0:
x2 + 3x − 18 = 0
(x + 6)(x − 3) = 0 ⇒ x = −6, 3
(e) 3x2 + 12x + 11 = 8:
3x2 + 12x + 11 = 2
3x2 + 12x + 9 = 0
x2 + 4x + 3 = 0
(x + 3)(x + 1) = 0 ⇒ x = −3, −1
The last one cannot be factored easily, so we use the quadratic equation.
(f) 3x2 + 12x + 11 = 0:
3x2 + 12x + 11 = 0 → a = 3, b = 12, c = 11
p
−12 ± 122 − 4(3)(11)
x =
2(3)
√
−12 ± 144 − 132
=
√6
−12 ± 12
=
,or
6√
√
12
12
= −2 +
, −2 −
6
6