MATH 135
INVERSE FUNCTIONS
1. For each of the following, either: say that the function is not invertible and
give an output y that has more than one corresponding input x, or find the
inverse of the function and give its domain and range.
i. f (x) = 2x + 1
y = 2x + 1
x = 2y + 1
x − 1 = 2y
1
2 (x − 1) = y
f −1 (x) =
1
(x − 1)
2
Domain: (−∞, ∞) (range of f )
Range: (−∞, ∞) (domain of f )
ii. g(x) = x2 − 5x + 6
Not invertible, as g is not one-to-one. If g −1 existed, then g −1 (0) would equal
both 2 and 3 and therefore would not be a function.
√
iii. h(x) = 1/ x + 1
√
y = 1/ √x + 1
x
√= 1/ y + 1
y + 1 = 1/x
y + 1 = (1/x)2
y = (1/x)2 − 1
h−1 (x) = (1/x)2 − 1
Domain: (0, ∞) (range of h)
Range: (1, ∞) (domain of h)
2. Answer the following questions about ex and its inverse.
i. Graph y = ex on the axes below. Is this function one-to-one? Yes.
Then ex is invertible. Call its inverse ln(x) (the natural logarithm of x).
ii. What is the domain of ln(x)?
The range of ex , which is (0, ∞).
iii. What is the range of ln(x)?
The domain of ex , which is (−∞, ∞).
iv. What is ln(1)?
Since e0 = 1, we switch input and output to see that ln(1) = 0.
v. Graph ln(x) on the axes to the right.
vi. As x approaches 0 from the right, what happens to ln(x)?
As x → 0 from the right, ln(x) → −∞.
4
2
−0.5
0.5
1
1.5
2
−2
−4
ex
y=x
ln(x)