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1720 6.1 Student Notes

CALCULUS II NOTES - Stewart 8th edition
Section 6.1 Inverse Functions
ˆ f : D → R where R = (−∞, ∞) and D is a subset of the real numbers (D ⊆ R) is
shorthand for f is a function from D into R. In calculus, D is usually an interval or union
of intervals (e.g., D = (0, ∞) or D = (−∞, 0) ∪ (0, ∞))
ˆ Each x in D (x ∈ D) is assigned f (x) in R, where f (x) is the image of x.
ˆ D is the domain of f . In this case R is the codomain. We will also write D = dom f .
ˆ The collection of all f (x)’s where x ∈ D is called the range of f . We will often write ran
f or R for the range.
ˆ As usual, when specifying a function by a formula we will assume the domain D to be
the largest set on which the formula yields a well defined real number unless indicated
Example 6.1-1: Give the domain and range for f (x) =
(x + 1)2
Definition: Suppose f is a function with domain D and range R. The inverse function (when it
exists) f −1 : R → D is the function so that f −1 (y) = x if and only if f (x) = y.
ˆ dom f −1 = ran f
ˆ ran f −1 = dom f
ˆ f (f −1 (y)) = y for any y ∈ ran f and f −1 (f (x)) = x for any x ∈ dom f (the so called
cancellation equations).
ˆ f −1 need not exist
Example 6.1-2: Let’s revisit f (x) =
from the last example.
(x + 1)2
This shows f −1 cannot exist whenever there are points x1 6= x2 where f (x1 ) = f (x2 ).
Definition: f : D → R is one-to-one if and only if f (x1 ) = f (x2 ) only when x1 = x2 . Equivalently,
x1 6= x2 implies f (x1 ) 6= f (x2 ).
A function has an inverse if and only if it is one-to-one.
Technical Note: In our definition, the inverse function f −1 is only required to be defined on
the range of f . With this convention there is no need for f to be surjective or “onto” to have an
inverse. This may conflict with conventions you may be learning in other classes.
Horizontal Line Test: A function is one-to-one if and only if every horizontal line intersects
the graph of f at most once.
If a function is not one-to-one, shrinking the domain may allow us to find an inverse to a part
of a function (used later in section 7.5 extensively). For instance, although f (x) = 3(x + 1)−2 is
not one-to-one on its domain x 6= −1, it is one-to-one when restricted to the interval (−1, ∞)
(or to the interval (−∞, −1)). Thus, an inverse exists for f on the domain x > −1.
Steps for Finding an Inverse Function
1. Check if f is one-to-one (if this is reasonable).
2. Set y = f (x) and solve for x in terms of y.
3. The result is x = f −1 (y).
4. It is “traditional” that x denote the input of a function. Swap x and y to obtain y = f −1 (x).
This is also done so that f and f −1 can be graphed on the same plane.
Example 6.1-3:
a) f (x) = 3x + 4. Find the inverse.
b) f (x) =
for x > −1. Find the inverse.
(x + 1)2
c) f (x) = − x − 2. Find the inverse and give the domain of the inverse.
If one graphs f and f −1 on the same plane, one sees that the graph of f −1 is the mirror image
of the graph of f with respect to the line y = x.
For instance, in examples (6.1-3a) and (6.1-3c) above, if we graph the f and f −1 in the same
plane, we find the following.
The Inverse Function Theorem: Suppose f : I → R is a continuous one-to-one function
where I is an interval. Then f −1 is continuous (on the range of f ). Moreover, if f is also
differentiable at a ∈ I and f 0 (a) 6= 0, then f −1 is differentiable at b = f (a) and (f −1 )0 (b) = 0 .
f (a)
Proof. We will not prove the full theorem; we will assume the continuity of f −1 (a proof of this
is in an appendix to the textbook). By definition of the derivative,
f −1 (y) − f −1 (b)
= lim
= 0
x→a f (x) − f (a)
f (a)
(f −1 )0 (b) = lim
where the second to last equality follows from the continuity of f −1 (that is, y → b implies
x → a). Example 6.1-4: Consider f (x) = x5 + 2x3 + x − 1. Find the slope of y = f −1 (x) at the point
(−1, 0).
Example 6.1-5: f (x) = (x3 + 1) . Find (f −1 )0 (8).