Examining the function

Question

I am having trouble with this function

$$f(x) = e^{2x}(2x-1)+1$$

I found the domain, domain is $\mathbb R$, it is neither even nor odd function.

I found that $f(x)=0 \Leftrightarrow x=0$, because it is obvious. I am having trouble with finding the sign of this function. That "+1" is making my life miserable. I can easily find the sign of $e^{2x}(2x-1)$, but what do I do with the constant?

Do I find the minimum of $e^{2x}(2x-1)$ and then add constant to it, and then check the sign?

That seems like too much work, considering that this function is actually a numerator of the first derivative of function that appeared in other graphing assignment. There has to be a better way...

The original graphing assignment was $$g(x)=\ln(e^{\frac{2x}{1-e^{2x}}})$$ You can ignore this one when writing the answer. I included it for the sake of context.

EDIT: By finding the "sign" of a function I mean finding the intervals for which $f(x)>0$ and $f(x)<0$.

Answer 1

If you know there is only one zero of the function, then you can easily find the sign of the function by finding the value of a single point of the function on either side of the zero. Since the function is continuous, it will never cross the x-axis again!

Answer 2

Since $f(x)$ is a continuous function, in order to find the intervals of positivity/negativity of $f(x)$ it is enough to understand where the zeroes of $f(x)$ lie. We have a little issue since $$ e^{2x}(2x-1)+1 = 0 $$ is a trascendental equation, but $x=0$ is clearly a root, and actually a double root, since $f'(0)=0$.
Due to $f'(x)=4x e^{2x}$, we have that $f$ is increasing on $\mathbb{R}^+$ and decreasing on $\mathbb{R}^-$. It follows that $f(x)$ is always positive on $\mathbb{R}\setminus\{0\}$.

Answer 3

What are the roots of $f(x)$? As $f$ is continuous its positive and negative intervals will be between the roots.

You say it is obvious that $f(x) = 0 \iff x = 0$. I that case as $f(-1) = \frac 1{e^2}(-3) + 1 > -\frac 34 + 1 > 0$ then for $x \in (-\infty, 0); f(x) > 0$. And as $f(1) > 0$ then for $x \in (0,\infty); f(x) > 0$.

But I don't see that it is obvious that $f(x) =0 \implies x=0$.

$e^{2x}(2x - 1) + 1 = 0 \implies$

$e^{2x} = -\frac 1{2x-1}$ As $e^{2x} > 0$ then $ x < 1/2$. For $x < 1/2$, $-\frac 1{2x-1}$ is decreasing and $e^{2x}$ is increasing so there is only one solution and it is $x=0$.

So we are done.