questions on functions

**wintersoltice** · August 7th 2011, 12:54 AM

Given that
f(x)= $\frac{5-x}{1-x}$

(a) Explain why f has an inverse and show that $f^{-1}=f$ .
(b) Evaluate $f^{51}(4)$ .

i can do part a,
but have no idea how to start for part b.

any help is appreciated. =)

**Siron** · August 7th 2011, 01:33 AM

I guess you've to calculate the 51th derivative of the function $f$ , I should try to recognize a pattern in you derivatives.

**HallsofIvy** · August 7th 2011, 03:06 AM

$f(f(x))= \frac{5- \frac{5- x}{1- x}}{1- \frac{5- x}{1- x}}$
Multiply both numerator and denominator by 1- x:
$f(f(x))= \frac{5(1- x)- (5- x)}{1- x- (5- x)}= \frac{ 5-5x-5+ x}{1- x-5+ x}= \frac{-4x}{-4}= x$

That shows both that f is invertible and that $f^{-1}(x)= f(x)$ .

**Plato** · August 7th 2011, 03:06 AM

Originally Posted by wintersoltice

Given that f(x)= $\frac{5-x}{1-x}$
(a) Explain why f has an inverse and show that $f^{-1}=f$ .

(b) Evaluate $f^{51}(4)$ .

I think that the notation $f^{51}$ refers to function composition.
For example: $f^5 (x) = f \circ f \circ f \circ f \circ f(x) = f\left( {f\left( {f\left( {f\left( {f(x)} \right)} \right)} \right)} \right) = ?$

**HallsofIvy** · August 7th 2011, 03:09 AM

Good point! And the fact that f is its own inverse makes that trivial!

**wintersoltice** · August 7th 2011, 04:00 AM

i found the solution from my school's online portal.

but i don't understand the steps. i mean from step 2 to 3.

step1 $f^{51} (4)$
step2 $=f[f^{50}(4)]$
step3 $=f(4)$ as $(# f^2(x)=f^{-1}f(x)=x)$
step4 $=\frac{5-4}{1-4}$
step5 $=\frac{-1}{3}$