Answer
If $c \gt \frac{1}{8},$ then the function $f(x) = cx+\frac{1}{x^2+3}$ is increasing on $(-\infty,\infty)$
Work Step by Step
$f(x) = cx+\frac{1}{x^2+3}$
$f'(x) = c+\frac{-2x}{(x^2+3)^2}$
If $f$ is increasing on $(-\infty,\infty)$, then $f'(x) \gt 0$ for all values of $x$
Let $g(x) = \frac{-2x}{(x^2+3)^2}$
To find the minimum value of $g(x)$, we can find $g'(x)$:
$g'(x) = \frac{(-2)(x^2+3)^2-(-2x)(2)(x^2+3)(2x)}{(x^2+3)^4}$
$g'(x) = \frac{(-2)(x^2+3)-(-2x)(2)(2x)}{(x^2+3)^3}$
$g'(x) = \frac{-2x^2-6+8x^2}{(x^2+3)^3}$
$g'(x) = \frac{6x^2-6}{(x^2+3)^3} = 0$
$6x^2-6=0$
$x^2-1 = 0$
$x =\pm 1$
$g(1) = \frac{-2(1)}{(1^2+3)^2} = -\frac{1}{8}$
$g(-1) = \frac{-2(-1)}{((-1)^2+3)^2} = \frac{1}{8}$
The function $g(x) = \frac{-2x}{(x^2+3)^2}$ has the minimum value of $-\frac{1}{8}$
Therefore:
$f'(x) = c+\frac{-2x}{(x^2+3)^2} \geq c-\frac{1}{8}$
If $c \gt \frac{1}{8},$ then $f'(x) \gt 0$ for all values of $x$
If $c \gt \frac{1}{8},$ then the function $f(x) = cx+\frac{1}{x^2+3}$ is increasing on $(-\infty,\infty)$
