functions

if f(x) =1 + ax (a≠0) is the inverse of itself then what is the value of a?

5 Answers

11
virang1 Jhaveri ·

y = 1 + ax
x = y - 1/a
y = x
therefore
1 +a x = x - 1 /a
a + a2x = x - 1
xa 2 +a - (x-1)
a = -1± √1 + 4x2 -4x/2x
a = -1 ±√(1-2x)2/2x
a = -1 + 1 - 2x/2x
a = -1
If in ± if we take '-' then a will be dependent on x but since a is a constant therefore '+' is taken

Another method is shown below

1
Shriya ·

the problem is that the answer is a constant (ie -1) not a function of x!!

11
virang1 Jhaveri ·

f(x)= 1 +a x
f-1(x) = x - 1 /a

Therefore
For n = 1
a = -1
For n = 2
a = -1 or 1/2
for n = 3
a = -1 or 2/3

Since a cannot be negative or the inverse will not be equal therefore a=-1 since it satisfies all values of a

1
Shriya ·

hey can you please define n ?!?!
f(x)=1+ax
let y=f-1(x)
then f(y)= x
1+ ay =x
y=(x-1)/a
f-1(x)=(x-1)/a
now f(x)=f-1(x)
so we have x=(1+a)/(1-a2) i am not getting what conclusion can we draw from this result ???

1
Shriya ·

thanks i got it

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