Strictly increasing on R

Choose correct options , more than one may be correct

Let f be the function defined by

$$h(x)=e^x (x-1)+x^2$$

we've :

1. $h$ is positive on $(0,\infty)$
2. $h$ is negative on $(0,1)$
3. $h$ is strictly increasing on $(0,\infty)$
4. The equation $h(x) = 0$ has a unique solution in $\mathbb{R}$

Graph

• From Calculator and unit converter of google search i'm going to say $(3)$, Indeed

In this proof i used Variation table which seems to be a very French thing. They are used to sum up the variations of a function.

The following diagram gives a variating table for the function h(x) defined on R

$\begin{array}{|c|cc|}\hline x & -\infty \qquad & 0 &\qquad +\infty \\ \hline x & - & 0 & + \\ \hline (e^x +2)& + & + & + \\ \hline h'(x)& - & 0 & + \\ \hline h(x)& \searrow &-1 & \nearrow \\ \hline \end{array}$

which means that $h$ is strictly increasing on $(0,\infty)$

• Would you please show me why others options aren't correct ?

For example There are several ways to prove that (4) is false we can show they are not applicable:

• Use the contraction principle (Banach fixed-point theorem ) to prove that the statement (the equation h(x)=0 has a unique solution) is false
• or use the Intermediate value theorem to prove it false

Any extra details is welcome.

• For the first and second ones you can argue by continuity by taking the limit at $0$ in the first one and at $1$ in the second one. you can always just evaluate the function at suitable values. – Git Gud Apr 8 '14 at 12:16

1) Clearly $h(0) = -1$. Since $h$ is continuous there exist values $x > 0$ with $h(x) < 0$.
2) Clearly $h(1) = 1$. Since $h$ is continuous there exist values $0 < x < 1$ with $h(x) > 0$
4) You can use the intermediate value theorem. Since $h(2) > 0$, $h(0) < 0$, and $h(-2) = \dfrac{-3}{e^2} + 4 > 0$, $h$ must have (at least) two real zeroes, one positive and one negative.
• Thanks would you please give the general rule ( h(a)=b and by continuity there exist .. ) which used in 1 and 2 did you use The Definition of Continuity $x - c | < \delta \Rightarrow | h(x) - h(c) | < \varepsilon. \,$ – Educ Apr 8 '14 at 13:03
• @Educ in both cases choose $\delta$ so that $|x-c| < \delta$ implies $|h(x) - f(c)| < 1$. In the case of 1) you have $|h(x) - h(0)| < 1$ implies $h(x) <h(0) + |h(x) - h(0)| < 0$. – Umberto P. Apr 8 '14 at 13:07
• @ Umberto P. i can't see for the second case, beside for (1) x>0 is produced from $|x| < \delta$(ball) but if i remember there is may be a fact that said if we've something true for ball it still holds for $\mathbb{R}$ sorry if i'm confused i forgot that stuff – Educ Apr 8 '14 at 13:26
• $|h(x) - h(1)| < 1$ implies $h(x) > h(1) - |h(x) - h(1)| > 0$. – Umberto P. Apr 8 '14 at 13:27
• The inequality above is valid in a neighborhood of $1$ that includes numbers $x$ in between $0$ and $1$. – Umberto P. Apr 8 '14 at 13:51