Strictly increasing on R

Question

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.

Answer 1

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.