Find the range of the function ##f(x) = \frac{e^{2x}-e^x+1} {e^{2x}+e^x+1}##

[tellmesomething]

Homework Statement

Title

Relevant Equations

Sridhacharya formulate: $\frac{-b+-√(b²-4ac)} {2a}$

My attempt:
Say f(x)=y=$\frac{e^{2x}-e^x+1} {e^{2x}+e^x+1}$
So I cant rearrange to write in terns of x
$e^{2x}(y-1)+e^x(y+1)+y-1,=0$
Now we know that e^x will not be defined for y=1
Therefore when y≠1
$e^x = \frac{ -y-1+-√((y+1)^2-4(y-1)^2)} {2(y-1)}$
So we can put an inequality on the discriminant, as it should be greater than equal to zero for e^x to be defined.

And from there we get that it will only be defined for y=[1/3,3]-{1}

So range would be y=[1/3,3] - {1}

But this is not the answer , please tell me where im going wrong

Correct answer is [1/3,1)

If we graph the function on desmos we can also see that this is the values y attains. I dont get where im going wrong...

 

[fresh_42]

Homework Statement: Title
Relevant Equations: Sridhacharya formulate: $\frac{-b+-√(b²-4ac)} {2a}$

But this is not the answer , please tell me where im going wrong

That would require to see what you have done. You left out too many steps.
I would start with $e^{2x} \pm e^x +1=e^{2x}\pm 2e^{x}+1\mp e^x=(e^x\pm 1)^2\mp e^x.$
Now you can rewrite $y=\dfrac{(e^x-1)^2+e^x}{(e^x+1)^2-e^x}$ and examine numerator and denominator.

 

[tellmesomething]

That would require to see what you have done. You left out too many steps.
I would start with $e^{2x} \pm e^x +1=e^{2x}\pm 2e^{x}+1\mp e^x=(e^x\pm 1)^2\mp e^x.$
Now you can rewrite $y=\dfrac{(e^x-1)^2+e^x}{(e^x+1)^2-e^x}$ and examine numerator and denominator.

But I didnt leave out many steps I just cross multiplied the function to get a quadratic in terms of e^x....

Also I get what you're suggesting but I posted my approach because I would like to know where $my$ logic went wrong..

 

[fresh_42]

Let me see in detail.
\begin{align*}
y&=\dfrac{e^{2x}-e^x+1}{e^{2x}+e^x+1}\\
y&\cdot (e^{2x}+e^x+1)=e^{2x}-e^x+1\\
0&=e^{2x}(y-1)+e^x(y+1) +(y-1)\\
\end{align*}
If $y=1$ then $-e^x=e^x$ which is impossible since $e^x\neq 0.$ So we may assume $y\neq 1.$
\begin{align*}
0&=e^{2x}+e^x\cdot \dfrac{y+1}{y-1} +1 \\
e^x &= -\dfrac{y+1}{2y-2}\pm \sqrt{\dfrac{(y+1)^2}{4(y-1)^2}-1}\\
&=\dfrac{y+1}{2-2y}\pm \sqrt{\dfrac{(y+1)^2-(4y^2-8y+4)}{4y^2-8y+4}}\\
&=\dfrac{y+1}{2-2y}\pm\sqrt{\dfrac{-3y^2+10y-3}{4y^2-8y+4}}\\
&=\dfrac{1}{2-2y}\left(y+1\pm \sqrt{-3y^2+10y-3}\right)\\
&=\dfrac{1}{2-2y}\left(y+1\pm \sqrt{(-3)\left(y^2-\dfrac{10}{3}y+\dfrac{25}{9}\right)-3+\dfrac{25}{3}}\right)\\
&=\dfrac{1}{2-2y}\left(y+1\pm \sqrt{(-3)\left(y-\dfrac{5}{3}\right)^2+\dfrac{16}{3}}\right)
\end{align*}

Now we have
\begin{align*}
0 &\leqq (-3)\left(y-\dfrac{5}{3}\right)^2+\dfrac{16}{3} \text{ and }\\
0&<\dfrac{1}{2-2y}\left(y+1\pm \sqrt{(-3)\left(y-\dfrac{5}{3}\right)^2+\dfrac{16}{3}}\right)
\end{align*}
This is what you left out and it isn't even the complete set of inequalities.

Since the first inequality is equivalent to $y\in \left[\dfrac{1}{3}\, , \,3\right]$ which is what you have, I assume that you have simply forgotten that $e^x > 0$ must hold, too.

 

[tellmesomething]

Let me see in detail.
\begin{align*}
y&=\dfrac{e^{2x}-e^x+1}{e^{2x}+e^x+1}\\
y&\cdot (e^{2x}+e^x+1)=e^{2x}-e^x+1\\
0&=e^{2x}(y-1)+e^x(y+1) +(y-1)\\
\end{align*}
If $y=1$ then $-e^x=e^x$ which is impossible since $e^x\neq 0.$ So we may assume $y\neq 1.$
\begin{align*}
0&=e^{2x}+e^x\cdot \dfrac{y+1}{y-1} +1 \\
e^x &= -\dfrac{y+1}{2y-2}\pm \sqrt{\dfrac{(y+1)^2}{4(y-1)^2}-1}\\
&=\dfrac{y+1}{2-2y}\pm \sqrt{\dfrac{(y+1)^2-(4y^2-8y+4)}{4y^2-8y+4}}\\
&=\dfrac{y+1}{2-2y}\pm\sqrt{\dfrac{-3y^2+10y-3}{4y^2-8y+4}}\\
&=\dfrac{1}{2-2y}\left(y+1\pm \sqrt{-3y^2+10y-3}\right)\\
&=\dfrac{1}{2-2y}\left(y+1\pm \sqrt{(-3)\left(y^2-\dfrac{10}{3}y+\dfrac{25}{9}\right)-3+\dfrac{25}{3}}\right)\\
&=\dfrac{1}{2-2y}\left(y+1\pm \sqrt{(-3)\left(y-\dfrac{5}{3}\right)^2+\dfrac{16}{3}}\right)
\end{align*}

Now we have
\begin{align*}
0 &\leqq (-3)\left(y-\dfrac{5}{3}\right)^2+\dfrac{16}{3} \text{ and }\\
0&<\dfrac{1}{2-2y}\left(y+1\pm \sqrt{(-3)\left(y-\dfrac{5}{3}\right)^2+\dfrac{16}{3}}\right)
\end{align*}
This is what you left out and it isn't even the complete set of inequalities.

Since the first inequality is equivalent to $y\in \left[\dfrac{1}{3}\, , \,3\right]$ which is what you have, I assume that you have simply forgotten that $e^x > 0$ must hold, too.

Oh damn! I really forgot that though the second inequality does look tedious to solve . Anyways Thankyou so much :-)

 

[fresh_42]

Oh damn! I really forgot that though the second inequality does look tedious to solve . Anyways Thankyou so much :-)

It's not too hard.
\begin{align*}
y\in \left[\dfrac{1}{3},3\right] &\Longrightarrow \dfrac{1}{2-2y}\in \left[-\dfrac{1}{4},\dfrac{3}{4}\right]\\
y\in \left[\dfrac{1}{3},3\right]&\Longrightarrow y+1\pm \sqrt{\ldots} \in \left[\dfrac{4}{3},4\right]\pm \left[\dfrac{1}{\sqrt{3}},\sqrt{3}\right]
\end{align*}
etc. Something like that. But I would check the result on WA to test possible mistakes.

 

[WWGD]

Well, you have obvious lower, upper bounds; 0,1, then you can use continuity/ mean value theorem.

 

[pasmith]

Simplify: \begin{split}<br /> \frac{e^{2x} - e^x + 1}{e^{2x} + e^x + 1} &amp;= 1 - \frac{2e^x}{e^{2x} + e^x + 1}\\<br /> &amp;= 1 - \frac{2}{2\cosh x + 1}.\end{split} Now the minimum is <br /> 1 - \frac{2}{2\cosh(0) + 1} = 1 - \frac23 = \frac13 and the supremum in the limit |x| \to \infty is 1. Thus <br /> \frac13 \leq \frac{e^{2x} - e^x + 1}{e^{2x} + e^x + 1} &lt; 1.

 

[SammyS]

@tellmesomething ,
I don't know how familiar you are with hyperbolic functions in general or with the cosh (hyperbolic cosine) function in particular.

Using basic operations for inequalities, the version of $f(x)$ given by @pasmith can be used to find the range of $f(x)$.

$\displaystyle \quad\quad\quad\quad f(x)=1 - \frac{2}{2\cosh(x) + 1}$

$\displaystyle \cosh(x) \ge 1$ for all real values of $x$ . Therefore,

$\displaystyle \quad\quad\quad\quad 3\le 2\cosh(x) + 1$ .

$\displaystyle \quad\quad\quad\quad\frac 1 3 \ge \frac 1 {2\cosh(x) + 1} \gt 0$ .

Multiply by $-2$ .

$\displaystyle \quad\quad\quad\quad\frac {-2} 3 \le \frac {-2} {2\cosh(x) + 1} \lt 0$

Adding $1$ gives

$\displaystyle \quad\quad\quad\quad\frac {1} 3 \le 1+\frac {-2} {2\cosh(x) + 1} \lt 1$

Finally: $\displaystyle \quad \frac {1} 3 \le f(x) \lt 1 \quad$ for all real values of $x$ .