is f(x) increasing or decreasing? on $(-1, 0]$

Question

Let the function $f(x)$ be defined by $f(x)= \sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{(2n+1)!}x^n$. Since f(x) is a function, then it can increase and decrease. So, in the interval $(-1, 0]$, is f(x) increasing or decreasing? Here we restrict x to any real number between this interval. Fortunately, I found from a previous question I asked that when x<0, $f(x)=\frac{\sinh( \sqrt{x} )}{\sqrt{x}}$. Credits to the people that answered that question. Graphing this function, I found that it is decreasing throughout. Of course, at $x=0$, the entire function is $0$, and it is neither decreasing or increasing.

Answer 1

You don't even need to find a closed form for $f$. If $x \in (-1, 0]$, then the series converges, and $|x| = -x$. We therefore have $$f'(x) = \sum_{n=0}^\infty \frac{n(-1)^{n - 1}}{(2n + 1)!}x^{n-1} = \sum_{n=0}^\infty \frac{n(-1)^{n - 1}}{(2n + 1)!}(-1)^{n-1}|x|^{n-1} = \sum_{n=0}^\infty \frac{n}{(2n + 1)!}|x|^{n-1} > 0,$$ for all $x \in (-1, 0]$ except $x = 0$. From this we can conclude via the MVT that $f$ is strictly increasing on this interval.