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I have this question:

For the function $f(x)=x-2\ln(x^2+3)$:

Find the two stationary points of this function, and enter them in the increasing order.

I know how to find the stationary points themselves, but I'm not sure how to get rid of the $\ln$ so that I could actually do that.

I thought it might be something like:

$f(x) = x-\frac{2}{x^2+3}$

but when I tried to find them with that it came up incorrect.

I was wondering if anybody might have any ideas?

Thanks!

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    $\begingroup$ What do you mean by stationary points? $\endgroup$
    – zz20s
    Jan 13, 2016 at 13:44
  • $\begingroup$ Isn't a stationary point when $f(x)=x$? If so it is easy to find the two solutions (if $x \in \mathbb{C}$). $\endgroup$
    – N74
    Jan 13, 2016 at 13:48
  • $\begingroup$ @nhz: Do you mean finding $f'(x) = 1-\frac{4 x}{x^2+3} = 0$ and finding $x = 1, 3$? $\endgroup$
    – Moo
    Jan 13, 2016 at 13:54

1 Answer 1

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I'm assuming that the stationary point is where the derivative of the function is equal to zero. Please correct me if this isn't true.

Anyway, you've calculated the derivative incorrectly.

For $f(x)=x-2\ln(x^2+3)$, $f'(x)=1-\frac{4x}{x^2+3}.$ Do you know the product or chain rules?

To find where this is equal to zero, we set $1-\frac{4x}{x^2+3}=0$

Thus, we get $1=\frac{4x}{x^2+3} \Rightarrow x^2+3=4x \Rightarrow 0=x^2-4x+3$.

Can you solve the quadratic?

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