f(x) = 1/x [(x+c)/(n+1)]^(n+1) , n, c are constants > 0. x > 0
I just want to clarify why I'm getting two different stat pts.
First one is when I differentiate it just as it is. So f'(x) = (-1/x^2) [(x+c)/(n+1)]^(n+1)] + 1/x (1/(n+1) [(x+c)/(n+1)]^n = 0
I get it simplified to be -c(x+c)^n = 0 hence x = -c.
.
But when I change the form first:
f(x) = 1/(n+1)^(n+1) * [(x+c)^(n+1)] /x
Differentiating using quotient, and letting f'(x) = 0, (n+1)(x+c)^n * x - (x+c)^(n+1) = 0
So (x+c)^n [ (n+1)x - (x+c)] = 0
(x+c)^n [ nx + x - x - c] = 0
x = -c, nx = c
x = c/n
Why am I getting an extra one here?????????????
I just want to clarify why I'm getting two different stat pts.
First one is when I differentiate it just as it is. So f'(x) = (-1/x^2) [(x+c)/(n+1)]^(n+1)] + 1/x (1/(n+1) [(x+c)/(n+1)]^n = 0
I get it simplified to be -c(x+c)^n = 0 hence x = -c.
.
But when I change the form first:
f(x) = 1/(n+1)^(n+1) * [(x+c)^(n+1)] /x
Differentiating using quotient, and letting f'(x) = 0, (n+1)(x+c)^n * x - (x+c)^(n+1) = 0
So (x+c)^n [ (n+1)x - (x+c)] = 0
(x+c)^n [ nx + x - x - c] = 0
x = -c, nx = c
x = c/n
Why am I getting an extra one here?????????????