Is this intuitive by graph ? Or its needs rigorous proof A double differentiable function $f : \mathbb{R} \to \mathbb{R}$ satisfy $f(x)f^{\prime \prime}(x) \ne 0 \ \forall x \in \mathbb{R}$. Can we conclude $f(x)f^{\prime \prime}(x) >0$ for all $\mathbb{R}$ , even though continuity of $f''(x)$ is not given. I thought of making a contradiction by graph supposing $f(x)>0$ and $f''(x) <0$ , I thought graph would need to cut x axis hence after that $f(x)<0$ ( contradiction) but I don't have a good calculus based proof on this fact. As continuity of $f''(x)$ not given is also a problem for all $\mathbb{R}$
 A: Since $f(x)f''(x)\ne 0$, it follows that for all real $x, f''(x)\gt 0$ or $f''(x)\lt 0$ for all real $x$. 
Because if there exist $x_1,x_2$ such that $f''(x_1)\gt 0$ and $f''(x_2)\lt 0$ then by Darboux theorem, there exists $c\in (x_1, x_2)$ such that $f''(c)=0$, which is a contradiction. (Why? Refer Note)
Similarly $f(x)\gt 0$ for all real $x$ or $f(x)\lt 0$ for all real $x$.
WLOG suppose that $f"(x)\gt 0$ for all $x$ and $f(x)\lt 0$ for all $x$, it follows that $f(x)f"(x)\lt 0$ for all $x$.  There exist $x'$ such that $f'(x')\ne 0$. 
Case 1: $f'(x')\lt 0$ 
By LMVT,
For $x\lt x', f'(x)\lt f'(x')$ 
So for $x\lt x', f(x)-f(x')\gt xf'(x')\implies f(x)\gt f(x')+xf'(x')\implies f(x)$ will be positive when $x$ is negative enough. This will contradict our assumption that $f(x)\lt 0$ 
Case 2: $f'(x')\gt 0$ 
Can you take it from here? 
Note: Define $g(x)=f'(x)$. Since $g$ is continuous on $[x_1,x_2]$, it must attain maximum on the interval.Let $c\in [x_1,x_2]$ be the point where maximum is attained. If $c=x_1$, then $g'(c)=f''(x_1)=\lim_{x\to x_1^+}\frac {f'(x)-f'(x_1)}{x-x_1}\le 0$, contradiction (Note that since $f'(x_1)$ is maximum, Numerator is not positive) and similarly $c\ne x_2$. So $c\in (x_1,x_2)$. Since $c$ is a point of extrema of $g$ on $[x_1,x_2]$, we must have $g'(c)=f''(c)=0$.
A: As you observe that $f(x)$ and $f^{\prime \prime }(x)$ can't change sign as the derivative has the darboux property Now we can assume that $f(x) > 0 \ \forall x \in \mathbb{R} $ as if it was $<0$ we can apply it to $-f$.
For the sake of contradiction assume $f^{\prime \prime}(x) < 0 \ \forall x \in \mathbb{R} $
Consider $c = f^{\prime}(0)$ if $c =0$ then $\exists \alpha>0$ such that $f^{\prime}(\alpha) < 0$(WHy?)
Or $f(x) - f(\alpha) < f^{\prime}(\alpha) (x- \alpha) \ \forall x > \alpha$ And take $x \to \infty$ to contradict that $f(x) >0$
If $c >0$ then $f(0) - f(x)>-f^{\prime}(0)x \ \forall x <0$ again take $x \to - \infty$ and contradict that $f(x) >0$
Now I think you can proceed
