To test if two functions are inverses, you just plug them into each other right? If not, how do you test this?

For two functions to be inverses, we need to satisfy the following properties:

First, we need to determine the two sets A and B such that:

f(x) is a bijective function f : A --> B

g(x) is a bijective function f : B --> A

With the property that f(g(x)) = x for all x in B, and g(f(x)) = x for all x in A

The term "bijective" means that the function is both "injective" and "surjective", or "1-to-1" and "onto". A function f(x) is "1-to-1" if every value y in the range there exists a unique x, such that f(x) = y. A function f(x) is "onto" if the co-domain is equal to the range. I.e. if f(x) is defined as f : A --> B, for every y in B there exists at least one x in A such that f(x) = y.

Examples:

f(x) = x^2 and g(x) = sqrt(x)

These are not inverses because f(x) is not injective. f(2) = f(-2), which is something that can't be true for inverse functions.

f(x) = x^3 and g(x) = x^(1/3)

These are inverses, because both functions are bijective and we have:

f(g(x)) = (x^(1/3))^3= x, and g(f(x)) = (x^3)^(1/3) = x

f(x) = tan(x) and g(x) = arctan(x)

These are also not inverses because tan(x) is not injective.

f(x) = x+1 and g(x) = x-1

These are invereses, because they're both bijective and:

f(g(x)) = (x-1)+1 = x, and g(f(x)) = (x+1)-1 = x

f(x) = e^x and g(x) = ln(x), where f : R --> R+ and g : R+ --> R

The additional specification is necessary, because otherwise f(x) would not be surjective. As is, both functions are bijections, and they satisfy:

f(g(x)) = e^(ln(x)) = x, and g(f(x)) = ln(e^x) = x

Note also that the functions f(x) = x and f(x) = k/x are equal to their own inverses. Kind of cool.

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@DukeZhang: Sorry, your example doesn't work. What if I take f(-3) = 9? The inverse will return 3, not -3. An inverse can't exist for f(x) = x^2 where f : R --> ({R+} + {0})

I don't mean to be harsh on you, it's just that this is one of the most common and annoying misconceptions about inverse functions.

log.base3(x+25) + 3 = log.base3(x+a million) + 5 log_3(x+25) - log_3(x+a million) = 5 - 3 log_3 ( (x+25) / (x+a million) ) = 2 3^2 = (x+25) / (x+a million) 9 * (x + a million) = x+25 9x + 9 = x + 25 9x - x = 25 - 9 8x = sixteen x = 2 y = log_3 ( 2 + 25 ) + 3 y = log_3 ( 27 ) + 3 y ? 6 intersection at (2 , 6) ============ y = log_5 (x+6) - 2 ====> change variables x = log_5 (y+6) - 2 ===> remedy for y x + 2 = log_5 (y+6) 5^(x + 2) = y + 6 5^(x + 2) - 6 = y = f^-a million(x)

yea, correct

e.g. Y1 = x^2 , Y2 = √x <= a basic inverse

lets select (3, 9) as a basic x, y value for Y1

you know if you plug 3 into Y1, you'll get 9

to test if Y2 is inverse, when you plug 9 into Y1, you should get 3 (and you will) hence Y2 is an inverse of Y1

the general rule is as follow, let (a, b) be any random pair of x,y on Y1

if Y2 is an inverse, then:

plug b into the the place of x, you should get a as the y value.