How do I show that this function is always $> 0$ 
Show that $$f(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +
 \frac{x^4}{4!} > 0 ~~~ \forall_x \in \mathbb{R}$$

I can show that the first 3 terms are $> 0$ for all $x$:
$(x+1)^2 + 1 > 0$
But, I'm having trouble with the last two terms. I tried to show that the following was true:
$\frac{x^3}{3!} \leq \frac{x^4}{4!}$
$4x^3 \leq x^4$
$4 \leq x$
which is not true for all $x$.
I tried taking the derivative and all that I could ascertain was that the the function became more and more increasing as $x \rightarrow \infty$ and became more and more decreasing as $x \rightarrow -\infty$, but I couldn't seem to prove that there were no roots to go with this property.
 A: $f$ is a polynomial, and therefore, is differentiable at all points.  Furthermore, as $x\to\infty$ or $x\to-\infty$, $f(x)\to+\infty$.  Thus, if $f(x)\le0$ for some $x$, then $f(x)\le0$ for some relative minimum.
$$f'(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}$$
$f'(x)=0$ for all relative minima.  However, if $f'(x)=0$, then $$f(x)=f'(x)+\frac{x^4}{4!}$$
Thus, $f(x)=\frac{x^4}{4!}>0$ for all relative minima $x\not=0$.  $x=0$ is not a relative minimum, because $f'(0)\not=0$, so this equation holds for all relative minima of $f$.  This contradicts our assumption, so $f(x)>0$ for all $x\in \mathbb R$.
A: Hint:
$$f(x) = \frac{1}{4} + \frac{(x + 3/2)^2}{3} +\frac{x^2(x+2)^2}{24}$$
A: You have had some good ideas so far. You tried to see when this was true: $$\frac{x^3}{3!} \leq \frac{x^4}{4!}.$$
You rearranged this to $4x^3\leq x^4$ but you made an incorrect conclusion when you divided by $x^3$ (if $x<0$ then the inequality sign should flip). Instead, lets divide by $x^2$ to get $4x \leq x^2$ or $x(x-4)\geq 0.$ This is true when $x\leq 0$ or $x\geq 4$ so the desired inequality is true in that range. 
For $0< x < 4$ we don't have $\frac{x^3}{3!} \leq \frac{x^4}{4!}$ but lets see if the other terms can save us. To do this, we need to see exactly how large $g(x) = x^3/3! - x^4/4!$ can be in $(0,4).$ We calculate that $g'(x) = -(x-3)x^2/6$ so $g$ increases when $0\leq x\leq 3$, the maximum occurs at $g(3)=9/8$, and then it decreases after that. 
This is good, because the $1+x+x^2/2$ terms obviously give at least $1$ from $x=0$, and will give us more as $x$ gets bigger. So we solve $1+x+x^2/2=9/8$ and we take the positive solution which is $\frac{\sqrt{5}-2}{2} \approx 0.118.$ So the inequality is definitely true for $x\geq 0.12$ because $g$ is at most $9/8$ and $1+x+x^2/2$ accounts for that amount in that range. 
Remember that $g$ was increasing between $x=0$ to $x=3$, so the largest $g$ can be in the remaining range is $g(0.12) = 873/315000 <1$, which is less than the amount $1+x+x^2/2$ gives us. So the inequality is also true for $0\leq x\leq 0.12$, so overall, for all $x.$
So all in all, the only trouble was for $x$ in $(0,4)$ and the contribution from the other terms was always enough to account for $x^3/3!$ when $x^4/4!$ wasn't enough.
A: Observe that
$$ e^x =  f(x) + \frac{x^5}{5!} + \cdots$$
Then show that for $x<0$ $$ f(x) > \frac{x^5}{5!} + \cdots$$
But $e^x > 0, \forall x\in R$
So, $$f(x) > e^x - f(x) \Rightarrow 2f(x)>e^x>0$$
