1. ## Functions

f(x) = 2 - (1/4)x^3 , x is a real number, x>0

Sketch the graph of f and find its range. Hence state the set of values of x for which ff^-1(x) = f^-1f(x)

[By f^-1(x) I mean f inverse. I am not sure of how to type out f inverse on the computer... ]

Thank You!

2. Originally Posted by Tangera

f(x) = 2 - (1/4)x^3 , x is a real number, x>0

Sketch the graph of f and find its range. Hence state the set of values of x for which ff^-1(x) = f^-1f(x)

[By f^-1(x) I mean f inverse. I am not sure of how to type out f inverse on the computer... ]

Thank You!
Provided you have a 1-to-1 function (which you do in this case), $f(f^{-1}(x)) = x\,$ over $\text{dom}f^{-1}$ and $\, f^{-1}(f(x)) = x\,$ over $\text{dom} f$.

So for $f(f^{-1}(x)) = f^{-1}(f(x))$ you need the set of values of x such that $\text{dom}f = \text{dom}f^{-1}$.

But you know that for any 1-to-1 function f, $\text{dom}f^{-1} = \text{ran} f$.

So you need the set of values of x such that $\text{dom}f = \text{ran}f$. From the graph you can see that these values will be $0 \leq x \leq 2$.

So $f(f^{-1}(x)) = f^{-1}(f(x)) = x$ for $0 \leq x \leq 2$.

3. erm...may I know what is dom f and ran f? Thank you!

4. Originally Posted by Tangera
erm...may I know what is dom f and ran f? Thank you!
It is shorthand:

dom f is the domain of the function f.

ran f is the range of the function f.

5. Ahhh...I see. Thank you!