Inverse functions flip the roles of input and output, allowing us to "undo" a function's effect. They're crucial for solving equations and understanding relationships between variables. We use tools like the horizontal line test to check if a function is invertible.
Inverse trigonometric functions help us find angles from ratios, with specific domains and ranges. When we combine a function with its inverse, we get the identity function. This relationship is key in calculus, especially when dealing with derivatives and solving complex equations.
Inverse Functions
Horizontal line test for invertibility
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Determines if a function is one-to-one meaning each x-value corresponds to exactly one y-value and vice versa
Functions that are not one-to-one fail the horizontal line test because some x-values map to multiple y-values (parabola)
Involves drawing horizontal lines across the graph of the function
If any horizontal line intersects the graph more than once, the function is not invertible (cubic function )
If no horizontal line intersects the graph more than once, the function is invertible (exponential function )
A function that passes the horizontal line test is injective
Computation and graphing of inverses
Steps to find the inverse of a function f ( x ) f(x) f ( x ) :
Replace f ( x ) f(x) f ( x ) with y y y
Interchange x x x and y y y variables
Solve the resulting equation for y y y
Replace y y y with f − 1 ( x ) f^{-1}(x) f − 1 ( x ) to denote the inverse function
The domain of f − 1 f^{-1} f − 1 is the range of f f f and the range of f − 1 f^{-1} f − 1 is the domain of f f f
For example, if f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 with domain [ 0 , ∞ ) [0, \infty) [ 0 , ∞ ) , then f − 1 ( x ) = x f^{-1}(x) = \sqrt{x} f − 1 ( x ) = x with domain [ 0 , ∞ ) [0, \infty) [ 0 , ∞ )
Graphing inverse functions involves reflecting the graph of f f f across the line y = x y = x y = x
If ( 2 , 4 ) (2, 4) ( 2 , 4 ) is a point on the graph of f f f , then ( 4 , 2 ) (4, 2) ( 4 , 2 ) is a point on the graph of f − 1 f^{-1} f − 1
Applications of inverse trigonometric functions
Inverse trigonometric functions allow you to find the angle given a trigonometric ratio
sin − 1 ( x ) \sin^{-1}(x) sin − 1 ( x ) or arcsin ( x ) \arcsin(x) arcsin ( x ) outputs the angle whose sine is x x x
cos − 1 ( x ) \cos^{-1}(x) cos − 1 ( x ) or arccos ( x ) \arccos(x) arccos ( x ) outputs the angle whose cosine is x x x
tan − 1 ( x ) \tan^{-1}(x) tan − 1 ( x ) or arctan ( x ) \arctan(x) arctan ( x ) outputs the angle whose tangent is x x x
The domain of inverse trigonometric functions is limited to the range of their corresponding trigonometric function
sin − 1 ( x ) \sin^{-1}(x) sin − 1 ( x ) has domain [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] and range [ − π 2 , π 2 ] [-\frac{\pi}{2}, \frac{\pi}{2}] [ − 2 π , 2 π ] (first and fourth quadrants)
cos − 1 ( x ) \cos^{-1}(x) cos − 1 ( x ) has domain [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] and range [ 0 , π ] [0, \pi] [ 0 , π ] (first and second quadrants)
tan − 1 ( x ) \tan^{-1}(x) tan − 1 ( x ) has domain ( − ∞ , ∞ ) (-\infty, \infty) ( − ∞ , ∞ ) and range ( − π 2 , π 2 ) (-\frac{\pi}{2}, \frac{\pi}{2}) ( − 2 π , 2 π ) (first and fourth quadrants)
Example calculation: If cos ( θ ) = − 3 2 \cos(\theta) = -\frac{\sqrt{3}}{2} cos ( θ ) = − 2 3 and θ \theta θ is in the third quadrant, then θ = cos − 1 ( − 3 2 ) + π \theta = \cos^{-1}(-\frac{\sqrt{3}}{2}) + \pi θ = cos − 1 ( − 2 3 ) + π
Functions vs their inverses
Composing a function with its inverse results in the identity function
( f ∘ f − 1 ) ( x ) = x (f \circ f^{-1})(x) = x ( f ∘ f − 1 ) ( x ) = x for all x x x in the domain of f − 1 f^{-1} f − 1 (plugging in f − 1 ( x ) f^{-1}(x) f − 1 ( x ) into f ( x ) f(x) f ( x ) returns x x x )
( f − 1 ∘ f ) ( x ) = x (f^{-1} \circ f)(x) = x ( f − 1 ∘ f ) ( x ) = x for all x x x in the domain of f f f (plugging in f ( x ) f(x) f ( x ) into f − 1 ( x ) f^{-1}(x) f − 1 ( x ) returns x x x )
The derivative of an inverse function is related to the derivative of the original function
If f f f is differentiable and invertible, then f − 1 f^{-1} f − 1 is also differentiable
( f − 1 ) ′ ( x ) = 1 f ′ ( f − 1 ( x ) ) (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} ( f − 1 ) ′ ( x ) = f ′ ( f − 1 ( x )) 1 (reciprocal of the derivative of f f f evaluated at f − 1 ( x ) f^{-1}(x) f − 1 ( x ) )
Equations with inverse functions
Inverse functions can be used to solve equations of the form f ( x ) = a f(x) = a f ( x ) = a
Apply f − 1 f^{-1} f − 1 to both sides of the equation: f − 1 ( f ( x ) ) = f − 1 ( a ) f^{-1}(f(x)) = f^{-1}(a) f − 1 ( f ( x )) = f − 1 ( a )
Simplify the left side using the property ( f − 1 ∘ f ) ( x ) = x (f^{-1} \circ f)(x) = x ( f − 1 ∘ f ) ( x ) = x : x = f − 1 ( a ) x = f^{-1}(a) x = f − 1 ( a )
Inverse trigonometric functions can be used to solve trigonometric equations
To solve cos ( x ) = − 2 2 \cos(x) = -\frac{\sqrt{2}}{2} cos ( x ) = − 2 2 , apply cos − 1 \cos^{-1} cos − 1 to both sides:
cos − 1 ( cos ( x ) ) = cos − 1 ( − 2 2 ) \cos^{-1}(\cos(x)) = \cos^{-1}(-\frac{\sqrt{2}}{2}) cos − 1 ( cos ( x )) = cos − 1 ( − 2 2 )
x = cos − 1 ( − 2 2 ) + 2 π n x = \cos^{-1}(-\frac{\sqrt{2}}{2}) + 2\pi n x = cos − 1 ( − 2 2 ) + 2 πn or x = − cos − 1 ( − 2 2 ) + 2 π n x = -\cos^{-1}(-\frac{\sqrt{2}}{2}) + 2\pi n x = − cos − 1 ( − 2 2 ) + 2 πn , where n n n is an integer
Function properties and invertibility
A function is surjective if every element in the codomain is mapped to by at least one element in the domain
A monotonic function is either entirely increasing or entirely decreasing, making it invertible
Function composition (f ∘ g) can be used to create new functions, which may or may not be invertible depending on the properties of f and g