help me to find the inverse of a function

[nmaths]

find the inverse function if f(x)=x^2 - 3x

[bobabot1]

f(x)^-1
i.e 1/(x^2-3x)

[zaza]

Welcome to the Forums


No no no. That is not the inverse of a function. That is 1/f.

Look up "completing the square" on Google.

[Glaysher]

f(x)= x² - 3x

To find inverse first complete the square to get it in form

f(x) = (x - a)² + b for some constants a and b

Then to find inverse write

y = (x - a)² + b

Swap the x's and y's

x = (y - a)² + b

Then rearrange to make y the subject

Then replace y with f^-1

[Glaysher]

Oh and to have an inverse you will need to restrict the domain as the function does not have an inverse if you take the domain to be the set of all real numbers. Perhaps you left out the domain in the question. The actual domain will affect the final answer.

[bobabot1]

Zaza, you're totally right. My bad. Man, school can't come sonner can it.

[eranga262154]

f(x)= x² - 3x

To find inverse first complete the square to get it in form

f(x) = (x - a)² + b for some constants a and b

Then to find inverse write

y = (x - a)² + b

Swap the x's and y's

x = (y - a)² + b

Then rearrange to make y the subject

Then replace y with f^-1

Exactly...

This is the way to find the inverse of a function.

If we get f(x)^-1, it gives that 1/f(x). It's not the invers of a function.

If you are fine with a cartesian plan, check that 1/f(x) is a inverse of f(x).