Given the definition of the function: $$ f(x) = e^x - e^{-x} + \frac{2x}{(x^2+1)^2} , x∈(-∞,0]$$

What's the monotonicity of this function in the given range? I tried to calculate the derivation to this function, and I got the result: $$ f'(x) = e^x + e^{-x} + \frac{-6x^4 -4x^2+2}{(x^2+1)^4}$$, the result of the derivation is still a Transcendental Function, and I cannot determine the positive and negative of the derivative function in the range(i.e. x∈(-∞,0]).

How to determine the positive and negative of the derivative function to determine the monotonicity of the f(x) here?


1 Answer 1


Let's take a look at


We will prove that

$$\frac{-6x^4-4x^2+2}{(x^2+1)^4} > -1$$

This gives us that $$(x^2+1)^4 > 6x^4+4x^2-2$$

$$x^8+4x^6+6x^4+4x^2+1 > 6x^4+4x^2-2$$

$$x^8+4x^6+3 = (x^4)^2 + 4(x^3)^2 + 3 > 0$$

which is true.

Thus, we have shown that the above inequality is true.

And it should be easy to see that $$e^x+e^{-x} \geq 1$$

since $e^{-x} \geq 1$.

Thus, combining the two inequalities we yield $$f'(x) = e^x+e^{-x}+\frac{-6x^4-4x^2+2}{(x^2+1)^4} > 0 = -1+1$$

Thus, the function is monotonically increasing.

  • $\begingroup$ Works also for $x>0$, notice that $e^x+e^{-x}=2\cosh x\geq2$ for all $x$, and there are only even exponents in the fraction. $\endgroup$ Dec 30, 2023 at 3:43
  • $\begingroup$ Yes, it works for all reals also because $f'(x)$ is even. $\endgroup$ Dec 30, 2023 at 3:46
  • $\begingroup$ Indeed. ${}{}{}{}$ $\endgroup$ Dec 30, 2023 at 3:47
  • $\begingroup$ @AaaLol_dude So, will get the conlusion that $$ f(x) <0 $$ when x ∈(−∞,0) and $$ f(x) ≥0 $$ when x∈[0, −∞). Then consider f(x) is the derivative function of function $$ h(x) = e^x + e^{-x} - \frac{1}{x^2+1} $$, when x ∈(−∞,0), h(x) monotonically decreasing and increasing otherwise. But, how did you come up with the assumption: $$ \frac{−6x^4−4x^2+2}{(x^2+1)^4} > -1 $$ ? It appears to me that the assumption should first be based on that we know f(x) will be monotonically increasing. $\endgroup$
    – xmh0511
    Dec 30, 2023 at 5:44
  • $\begingroup$ If you compared the degree of each polynomial, you can see that it is a 4 / 8, meaning that the polynomial will get quite small. Moreover, the negative sign in front suggests a negative value, and so it would be worth a try. $\endgroup$ Dec 30, 2023 at 8:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .