Find the value of X which will produce critical values. y = ln ( x^2 + 1 ) I\'m

Question

Find the value of X which will produce critical values.
y = ln ( x^2 + 1 )
I'm guessing you find the derivative first and after make that equation equal to 0.
The derivative is going to be (1 / x^2 + 1)(2x) and I make that equal to 0. I'm slightly confused on what do after and how do I know if it's a critical value or not. Find the value of X which will produce critical values.
y = ln ( x^2 + 1 )
I'm guessing you find the derivative first and after make that equation equal to 0.
The derivative is going to be (1 / x^2 + 1)(2x) and I make that equal to 0. I'm slightly confused on what do after and how do I know if it's a critical value or not.
y = ln ( x^2 + 1 )
I'm guessing you find the derivative first and after make that equation equal to 0.
The derivative is going to be (1 / x^2 + 1)(2x) and I make that equal to 0. I'm slightly confused on what do after and how do I know if it's a critical value or not.

Explanation / Answer

y = ln ( x^2 + 1 )

y'= 1/(x^2 +1) x d/dx(x^2) = 2x/x^2+1

complete the square of (x^2 + 1 + 2x - 2x) = (x+1)^2 - 2x

this implies, 2x/ (x+1)^2 - 2x =0

it means, 2x=0 ==== x=0 is the critical point.

(x+1)^2 - 2x ===== x = -1 is the critical point

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