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Let \(f(x) = \frac{ax}{x + 1}.\)
Find the constant \(a\) so that \((f(x)) = x \) for all \(x \neq -1\)

 

I found that a=x+1 (im not sure if that's right but the question wants a constant for a. Help!

 

I really would appreciate it guys ;)

 Oct 29, 2019
 #1
avatar+109202 
+1

f(x)  = x

 

ax  = (x + 1) x

 

ax  = x^2 + x

 

x^2 + ( 1 - a)x  = 0

 

x ( x + (1 - a) )  =  0       ⇒   x  = 0    (reject)   or

 

x + (1 - a)  =  0

 

x  = - (1 - a)

 

x   =  a - 1

 

x + 1  =  a

 

This must be correct because

 

 ax              (x + 1)x

____  =     _______   =      x

x + 1            x + 1

 

 

 

cool cool cool

 Oct 29, 2019

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