If f(x)=x^2-2 and g(x)=1/(3x) (x does not equal 0), find f^-1(x) where x </ 0
? i don't get what u r asking
Uh...that makes two of us. If I got the problem I wouldn't be asking it.
See similar post:
http://www.jiskha.com/display.cgi?id=1248753911
If the post doesn't answer your question, please start a new thread.
To find the inverse of the function f(x) = x^2 - 2, we follow these steps:
Step 1: Replace f(x) with y.
y = x^2 - 2
Step 2: Swap x and y.
x = y^2 - 2
Step 3: Solve the equation for y.
x + 2 = y^2
Step 4: Take the square root of both sides.
±√(x + 2) = y
Since we are interested in finding f^-1(x) where x </ 0, we select the negative square root to get rid of the ± sign:
y = -√(x + 2)
Therefore, the inverse function of f(x) = x^2 - 2 is f^-1(x) = -√(x + 2), where x </ 0.