Functions question

On page 27 of the Pearson Edexcel C3 textbook example 13c says "find values of x such that f(x)=f^-1(x)" how did they simplify that to:

f(x) = f^-1(x)
f(x) = x

How can f^-1(x) just become x?

Reply 1

Notnek

Original post by mp_x

On page 27 of the Pearson Edexcel C3 textbook example 13c says "find values of x such that f(x)=f^-1(x)" how did they simplify that to:

f(x) = f^-1(x)
f(x) = x

How can f^-1(x) just become x?

f(x) = x is a function that satisfies f(x) = f^-1(x).

http://www.thestudentroom.co.uk/showthread.php?t=3540817

Try reading my post in this thread.

Or thinking of it graphically, the inverse fuction is a reflection of the function in the line y = x. So only the function y = x could be the inverse of itself.

(edited 8 years ago)

Reply 2

Zacken

Original post by mp_x

On page 27 of the Pearson Edexcel C3 textbook example 13c says "find values of x such that f(x)=f^-1(x)" how did they simplify that to:

f(x) = f^-1(x)
f(x) = x

How can f^-1(x) just become x?

You know that

f(x)

and

f^{-1} (x)

are simply reflections of one another in the line

y=x

. So when they give you the equation

f(x) = f^{-1}(x)

you know that the solutions to this equation lies on the line

y=x

allowing you to simplify your equation as

f(x) = x = f^{-1}x

.

Examples:

e^x = \ln x

, this one is hard to solve by itself - so let's use the fact that they are inverse functions of one another and satisfy

e^x = x = \ln x

which makes life slightly easier (although not very much so, unless you want to use the Lambert W function).

Here's another one, solve:

x^2 = \sqrt{x}

for

x>0

- they are inverses of one another and hence satisfy

x^2 = x

x^2 = x \iff x^2 - x = 0 \iff x(x-1) = 0

Reply 3

mp_x

Original post by notnek

f(x) = x is the only function that satisfies f(x) = f^-1(x).

http://www.thestudentroom.co.uk/showthread.php?t=3540817

Try reading my post in this thread.

Or thinking of it graphically, the inverse fuction is a reflection of the function in the line y = x. So only the function y = x could be the inverse of itself.

Original post by Zacken

You know that

f(x)

and

f^{-1} (x)

are simply reflections of one another in the line

y=x

. So when they give you the equation

f(x) = f^{-1}(x)

you know that the solutions to this equation lies on the line

y=x

allowing you to simplify your equation as

f(x) = x = f^{-1}x

.

Examples:

e^x = \ln x

, this one is hard to solve by itself - so let's use the fact that they are inverse functions of one another and satisfy

e^x = x = \ln x

which makes life slightly easier (although not very much so, unless you want to use the Lambert W function).

Here's another one, solve:

x^2 = \sqrt{x}

for

x>0

- they are inverses of one another and hence satisfy

x^2 = x

x^2 = x \iff x^2 - x = 0 \iff x(x-1) = 0

I'm still confused sorry :frown:

I get that f^-1f(x) = x because you're basically doing something then undoing it but I just don't get how f^-1(x) = x

Reply 4

mp_x

Original post by Zacken

You know that

f(x)

and

f^{-1} (x)

are simply reflections of one another in the line

y=x

. So when they give you the equation

f(x) = f^{-1}(x)

you know that the solutions to this equation lies on the line

y=x

allowing you to simplify your equation as

f(x) = x = f^{-1}x

.

Examples:

e^x = \ln x

, this one is hard to solve by itself - so let's use the fact that they are inverse functions of one another and satisfy

e^x = x = \ln x

which makes life slightly easier (although not very much so, unless you want to use the Lambert W function).

Here's another one, solve:

x^2 = \sqrt{x}

for

x>0

- they are inverses of one another and hence satisfy

x^2 = x

x^2 = x \iff x^2 - x = 0 \iff x(x-1) = 0

I now understand this graphically but not in principle as it doesn't follow the principle in my reply above

Reply 5

Notnek

Original post by mp_x

I'm still confused sorry :frown:

I get that f^-1f(x) = x because you're basically doing something then undoing it but I just don't get how f^-1(x) = x

The inverse of a function

f(x)

maps the output of the function back to the input.

So if

f(a) = b

then

f^{-1}(b) = a

.

Try it with

f(x) = x + 1

and

f^{-1}(x) = x - 1

f(a) = a + 1

f^{-1}(a + 1) = a

Now if

f = f^{-1}

then it must follow that if

f(a) = b

then

f(b) = a

. So this function must be able to "undo" its mapping.

Let's try a function

f(x) = x+1

. This clearly won't work because the mapping will always add 1 to the input.

A function can undo it's mapping if it maps any element to itself i.e. the function