Understanding Exponential Functions and Inverse Functions

1.
a)
h(t) = 5t + 7
h(t) is basically the function 5t shifted up by 7 units. It is therefore essentially an
exponential function. Exponential functions accept all real numbers as input, there-
fore domain of h(t) is:
Domain: (−∞, +∞)
Exponential functions decay and tend to zero as t → −∞. Since 7 is added to the
exponential term, it will tend to 7 as t → −∞. Exponential functions tend to ∞ as
they grow. Therefore, range of h(t) is:
Range: (7, +∞)
b)
r(m) = 2 + √3 − m
Square root of negative numbers is undefined, so 3−m cannot become less than zero.
3 − m = 0 =⇒ m = 3, for m > 3, r(m) is undefined, so the function can accept
numbers less than 3.
Domain: [3, −∞)
As m → −∞, r(m) tends to +∞. The least value r(m) achieves when the square
root term becomes zero, at which its value is 2.
Range: [2, +∞)
2.
f (x) = 3
2x + 1
a) Inverse
Step 1: y = 3
2x + 1
Step 2: Interchange x and y
x = 3
2y + 1

Step 3: Solve for y
2y + 1 = 3
x =⇒ y = 1
2
( 3
x − 1
)
Step 4: y = f −1(x)
f −1(x) = 1
2
( 3
x − 1
)
Step 5: Verify: (f −1 o f )(x) = x
(f −1 o f ) = 1
2
( 3
f (x) − 1
)
= 1
2
(
3
3
2x+1
− 1
)
= 1
2 (2x + 1 − 1) = x =⇒ Verified
Therefore, inverse of f (x) is f −1(x) = 1
2
( 3
x − 1
)
b)
f (x) accepts all values of x except the one for which 2x + 1 = 0, because at this
point value of f (x) is undefined. This point is x = −1
2 . Therefore,
Domain:
(
−∞, −1
2
)
∪
(
−1
2 , +∞
)
f (x) attains all values possible except 0 because the function tends to 0 as x gets
bigger and bigger but never actually reaches it. f (x) = 0 is an horizontal assymptote.
Range: (−∞, 0) ∪ (0, +∞)
c)
For an inverse function, the domain is just the range of f (x) and the range is domain
of f (x). Therefore,
Domain: (−∞, 0) ∪ (0, +∞)
Range:
(
−∞, −1
2
)
∪
(
−1
2, +∞
)

3.
f (x) = 3x2 + 1
a)
f (1) is the value of f (x) at x = 1. Therefore,
f (1) = 3(1)2 + 1 = 3 + 1 = 4
b)
f (1 + h) = 3(1 + h)2 + 1 = 3(1 + 2h + h2) + 1 = 3h2 + 6h + 4
c)
f (1) − f (1 + h)
h = 4 − (3h2 + 6h + 4)
h
= −3h2 − 6h
h
= −3h − 6
4.
x2 + y2 − 4x + 8y + 16 = 0
We want to reduce this equation into standard form of circle equation: (x − h)2 +
(y − k)2 = a2, where (h, k) is the circle center while a is the radius.

0 = x2 + y2 − 4x + 8y + 16
0 = x2 − 4x + y2 + 8y + 16
[Split 16 as 4+12]
0 = (x2 − 4x + 4) + y2 + 8y + 12
[Add 4 to both sides of the equation]
4 = (x2 − 4x + 4) + (y2 + 8y + 16)
[Terms in brackets are squares]
22 = (x − 2)2 + (y + 4)2
Comparing with standard form, we get: a = 2, h = 2 and k = −4. Therefore, the
circle has a center at (2, −4) and has a radius=2.
5.
a)
10 − 6x ≤ 14 − 4x
[Add 6x to both sides]
10 ≤ 14 + 2x
14 + 2x ≥ 10
[Subtract 14 from both sides]
2x ≥ −4
[Divide throughout by 2]
x ≥ −2
Therefore, solution to the inequality is:
x ∈ [−2, +∞)