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Please , can you verify only if the exercise is right?

We have the function with domain and range in $\mathbb{R}$, $f(x)=mx-ln(x^2+1)$. Determine the values of "$m$" so that the function is decreasing on $\mathbb{R}$.

The derivative of $f(x)=m-(2x)/(x^2+1)$

In order to make the function decrease on $\mathbb{R}$ , the derivative should be less than zero for every $x$ that belongs to $\mathbb{R}$.

$\dfrac{m-(2x)}{(x^2+1)}<0$ $mx^2-2x+m<0$

So $m>0$ and the discriminant should be less than zero

$D=4-4m^2<0$ $m$ belongs to $(-\infty; -1)$ in reunion with $(1,\infty)$

Is it right?

Hawk
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wonderingdev
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1 Answers1

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(Almost) as in your solution, we want to find $m$ such that $m-\frac{2x}{x^2+1}\le 0$ for all $x$.

Using the derivative of $\frac{2x}{1+x^2}$, we can show that $\frac{2x}{1+x^2}$ reaches an absolute minimum of $-1$ at $x=-1$. So, for any $m$, $m-\frac{2x}{x^2+1}$ reaches an absolute maximum at $x=-1$.

Thus we want $m+1\le 0$, giving $m\le -1$.

Remark: Alternately, we can, as you did, work with $mx^2-2x+m$. To make this $\le 0$ for all $x$, we clearly need $m\lt 0$. Then a discriminant analysis will work. It is safer to let $m=-k$. Then our condition becomes $kx^2+2x+k\ge 0$ for all $x$.

André Nicolas
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