# Some doubts with the sign of a derivative

Good evening to everyone. The derivative is defined in the following order: $$\frac{d}{dx} f(x)=\frac{{-x^2-x+11}}{\left(x+3\right)^2}e^{2-x}\:$$ for $x < -3$ $$\:\frac{d}{dx}f(x)=\frac{{x^2+x-11}}{\left(x+3\right)^2}e^{2-x}$$ for $-3< x< 2$ and $$\frac{d}{dx}f(x) = \frac{{x^2+x+1}}{\left(x+3\right)^2}e^{x-2}$$ for $x>2$. If you'll do the sign of the first one it'll be $${x^2+x-11} < 0$$ therefore only $$x_1 = \frac{-1+\sqrt{45}}{2}$$ verifies the conditions, for the second one it is $${x^2+x-11} > 0$$ therefore $x_1 = \frac{-1+\sqrt{45}}{2}$ and $x_1 = \frac{-1-\sqrt{45}}{2}$ , the last one $${x^2+x+1} > 0$$ has the solutions $x_1= \frac{-1-i\sqrt{3}}{2}$ and $x_2 = \frac{-1+i\sqrt{3}}{2}$(here I think that I did a mistake). Therefore our function should increase on the interval $(-\infty, \frac{-1-\sqrt{45}}{2})$ then decrease on the interval $(\frac{-1-\sqrt{45}}{2},\frac{-1+\sqrt{45}}{2})$ then increase again on the interval $(\frac{-1+\sqrt{45}}{2}, \frac{-1-i\sqrt{3}}{2})$ and decrease again on $(\frac{-1+i\sqrt{3}}{2},\infty)$. But Instead the solutions of the third equation are $x_1 = \frac{-1-\sqrt{5}}{2}$ and $x_2 = \frac{-1+\sqrt{5}}{2}$. And it is decreasing on $(-\infty, \frac{-1-\sqrt{45}}{2})$, then increasing on $(\frac{-1-\sqrt{45}}{2}, -3)$ then decreasing on $(-3,2)$ then increasing again on $(2,\infty)$. I don't get where I'm wrong.. neither with the results of quadratic equation nor with the intervals of monotony(sign) of the derivative.

• I will assume the given derivatives are the intended ones. The easiest is the interval $(2,\infty)$, for the derivative is defined everywhere on that interval, and is positive, so the function is increasing. Now let us deal with $x\lt -3$. Again, the derivative is defined everywhere. It is negative for $x\lt (-1-\sqrt{45})/2$, and positive in the interval $(-1-\sqrt{45})/2,3)$, so decreasing in $(-\infty, (-1-\sqrt{45})/2$ and increasing in $(-1-\sqrt{45})/2,3)$. The last interval is dealt with similarly. – André Nicolas Jun 27 '16 at 20:10
• If the solution sheet says that the solutions to the third equation are $(-1\pm\sqrt{5})/2$, either the solution sheet has a mistake or the numerator is supposed to be $x^2+x-1$. But it does not matter for the increasing/decreasing part, since these two roots are not in $(2,\infty)$. If the numerator is $x^2+x+1$, as in the post, the two roots are non-real, and so again we conclude the derivative is positive in $(2,\infty)$. – André Nicolas Jun 27 '16 at 20:26
• @Andre Nicolas thanks I understood perfectly!:) – T4yl0r Jun 27 '16 at 20:43
• You are welcome. You ran into difficulties because you tried to treat the three intervals simultaneously. – André Nicolas Jun 27 '16 at 20:45

## 1 Answer

We have a root for the first derivative at around $x\approx-3.854$ the other is negligible as the derivative is not defined with the particular function at that point.

With $x<-\frac{1+3\sqrt{5}}{2}, f'(x)<0$. Thus, $f(x)$ is decreasing for $x\in(-\infty,-\frac{1+3\sqrt{5}}{2})$ and increasing for $x\in( -\frac{1+3\sqrt{5}}{2},-3)$. (I substituted a random value for $x<-\frac{1+3\sqrt{5}}{2}$ and found that $f'(x)<0$)

For $-3<x<2$, the derivative defined as the same roots so now the root around $x\approx2.854$ is relevant and the other is negligible.

With $-3<x<2, f'(x)$ is negative. Thus, $f(x)$ is decreasing from $x\in(-3,2)$.

For $x>2$ the derivative defined does not have any real roots. With a random check, we find that at any point $x>2, f'(x)>0$. Thus, $f(x)$ is always increasing on $x\in(2,\infty)$.

• Can I ask how did you draw the conclusion that for $x < -\frac{1+3\sqrt{5}}{2}$ $\frac{d}{dx}\left(f\left(x\right)\right) < 0$ because my calculations give me $x$ belongs to $$( -\frac{1+3\sqrt{5}}{2}, -\frac{1-3\sqrt{5}}{2})$$ so $\frac{d}{dx}\left(f\left(x\right)\right) <0$ only on this interval so it says nothing about $(-\infty, -\frac{1+3\sqrt{5}}{2})$ – T4yl0r Jun 27 '16 at 20:26
• I'm not sure about the source of confusion. Try $x=-10. f'(-10)<0$. $f'(x)$ as a root at two values with the lower value at $x=-\frac{1+3\sqrt{5}}{2}$. So, $f'(x)<0$ for $x\in(-\infty,-\frac{1+3\sqrt{5}}{2})$. Then, this particular $f'(x)$ is not defined for $x\geq -3$ so for $x\in(-\frac{1+3\sqrt{5}}{2},-3), f'(x)>0$ – Melinda Jun 27 '16 at 20:32