Determine whether $f(x)$ is increasing or decreasing Let $f(x) = -x + (x^3/3!) + \sin(x)$
How do I determine if $f(x)$ is increasing or decreasing?
I have already found the derivative of this function which is:
$f'(x) = -1 + (x^2/2) + \cos(x)$
And I know I have to solve the inequalities for:
$-1 + (x^2/2) + \cos(x) \geq 0$ (increasing)
and
$-1 + (x^2/2) + \cos(x) \leq 0$ (decreasing)
But I am having some problems to solve them.
How should I proceed?
 A: Since $-1\leq \cos(x)\leq 1$, you have $-2\leq \cos(x)-1\leq 0$.  Since $\frac{x^2}{2}\geq 2$ for $x\geq 2$ and $x\leq -2$, your function will increase on those intervals.  So, you need to only consider $-2\leq x\leq 2$.  Both $\cos(x)$ and $\frac{x^2}{2}$ are even functions.  This allows us to only consider $-2\leq x\leq 0$.  On this interval $\cos(x)$ is an increasing function.  Thus $\cos(x)-1$ is also increasing.  In addition $\frac{x^2}{2}$ is increasing.  Since $-1+\cos(-2)+\frac{(-2)^2}{2}> 0$, the derivative must be greater than $0$ for all of $-2\leq x<0$.
A: As $f$ is odd, we need to study only the case $x\ge0$, if it is increasing, the same will be valid for corresponding regions in the negative axis also. 
Now we have $\cos x \le 1 \implies 1-\cos x \ge 0$. Integrating this from $0$ to $x$ successively will give you the result you seek... i.e.
$$\implies \int_0^x (1-\cos x) dx = x -\sin x \ge 0$$
$$\implies \int_0^x(x-\sin x) dx = \frac{x^2}2+\cos x -1\ge 0$$
A: You might want to think of finding the roots of the derivative and then determining if the function is positive of negative to the left and right of the roots. 
$$f'(x) = −1+(x^2/2)+\cos(x)$$
Use Wolfram Alpha to find the roots and see when it is positive and negative.
A: I would go with factoring using a sub-multiple trig ratio as it leads to verifying signs of $\sin(x)\over{x}$ to solve the inequalities which seems pretty natural to me
$-1+\frac{x^2}{2}+cos(x) = \frac{x^2}{2}-2sin^2(\frac{x}{2}) \ge 0$
$\implies x^2 - 4sin^2(\frac{x}{2}) \ge 0$
$\implies (x-2sin(\frac{x}{2}))(x+2sin(\frac{x}{2})) \ge 0$
A: I consider first the case for $x\geq 0$.
At $x=0$, we have  $f'(x)=−1+(x^2/2)+cos(x)=0$ and $f''(x)= x-sin(x)$ which at $x=0$ is also $0$. But note that for all $x \geq 0$ we have $f'''(x)= 1-cos(x)\geq 0$. So $f''(x)$ is increasing in x. Since $f''(0)=0$, it follows that $f''(x)>0$ for all $x>0$. This in turn, implies that $−1+(x^2/2)+cos(x)>0$ for all $x>0$ since $f'(0)=0$.
Since, $f'(x)=f'(-x)$ this establishes that the derivative is always (weakly) positive. You can strengthen it to always strictly positive for all $x\neq 0$.
