Question: User: find to domain and range
Assistant: 1. **State the problem.**
We need the domain of $y=\frac{\sqrt{6x^2+11x-7}}{\sqrt{|2x-5|+4x^2-2x-13}}$.
Because there are square roots in both the numerator and denominator, we must make sure:
- the expression inside the numerator square root is at least $0$,
- the expression inside the denominator square root is greater than $0$,
- and the denominator is not $0$.
2. **Set up the domain conditions.**
For the numerator:
$6x^2+11x-7\ge 0$
For the denominator:
$|2x-5|+4x^2-2x-13>0$
3. **Solve the numerator inequality.**
Factor the quadratic:
$6x^2+11x-7=(6x-3)(x+\tfrac{7}{3})$
A cleaner factorization is:
$6x^2+11x-7=(3x+7)(2x-1)$
So we solve:
$(3x+7)(2x-1)\ge 0$
The critical points are:
$x=-\tfrac{7}{3},\; x=\tfrac{1}{2}$
Since the parabola opens upward, the product is nonnegative outside the roots:
$x\le -\tfrac{7}{3}\quad \text{or}\quad x\ge \tfrac{1}{2}$
4. **Simplify the denominator expression by cases.**
We need:
$|2x-5|+4x^2-2x-13>0$
Use the absolute value split at $x=\tfrac{5}{2}$.
**Case 1: $x\ge \tfrac{5}{2}$**
Then $|2x-5|=2x-5$, so the denominator becomes:
$2x-5+4x^2-2x-13=4x^2-18$
So we need:
$4x^2-18>0$
$4x^2>18$
$x^2>\tfrac{9}{2}$
Since this case already has $x\ge \tfrac{5}{2}$, the condition becomes:
$x>\tfrac{3}{\sqrt{2}}$
**Case 2: $x<\tfrac{5}{2}$**
Then $|2x-5|=5-2x$, so the denominator becomes:
$5-2x+4x^2-2x-13=4x^2-4x-8$
Now solve:
$4x^2-4x-8>0$
Divide by $4$:
$\frac{\cancel{4}x^2-\cancel{4}x-\cancel{8}}{\cancel{4}}>0$
$x^2-x-2>0$
Factor:
$(x-2)(x+1)>0$
This is positive when:
$x<-1\quad \text{or}\quad x>2$
Because we are in the case $x<\tfrac{5}{2}$, both parts are allowed here.
So the denominator is positive for:
$x<-1\quad \text{or}\quad 2\tfrac{3}{\sqrt{2}}$
But notice $x>\tfrac{3}{\sqrt{2}}$ is already contained in $x>2$, so the combined denominator condition is simply:
$x<-1\quad \text{or}\quad x>2$
5. **Intersect the numerator and denominator conditions.**
Numerator condition:
$x\le -\tfrac{7}{3}\quad \text{or}\quad x\ge \tfrac{1}{2}$
Denominator condition:
$x<-1\quad \text{or}\quad x>2$
Now take the overlap:
- For the left side, $x\le -\tfrac{7}{3}$ already fits inside $x<-1$.
- For the right side, $x\ge \tfrac{1}{2}$ and $x>2$ together give $x>2$.
So the domain is:
$(-\infty,-\tfrac{7}{3}]\cup(2,\infty)$
6. **Final answer.**
The domain is $(-\infty,-\tfrac{7}{3}]\cup(2,\infty)$.
User: what about the range
1. **State the problem.**
We are finding the range of $y=\frac{\sqrt{6x^2+11x-7}}{\sqrt{|2x-5|+4x^2-2x-13}}$.
From the domain work, the function is defined for $x\in(-\infty,-\tfrac{7}{3}]\cup(2,\infty)$.
2. **Use the main range formula idea.**
For a function with square roots in a fraction, the output is always $y\ge 0$ because both numerator and denominator are nonnegative, and the denominator is positive on the domain.
So we now check whether $y$ can become $0$ and whether it can grow without bound.
3. **Check when the function can be $0$.**
A fraction with a square root numerator is $0$ when the numerator is $0$.
Set the numerator equal to $0$:
$$6x^2+11x-7=0$$
Factor:
$$6x^2+11x-7=(3x+7)(2x-1)$$
So the zeros are:
$$x=-\tfrac{7}{3},\quad x=\tfrac{1}{2}$$
Now test these in the domain.
- $x=-\tfrac{7}{3}$ is in the domain.
- $x=\tfrac{1}{2}$ is not in the domain.
At $x=-\tfrac{7}{3}$, the numerator is $0$ and the denominator is positive, so the function value is $0$.
4. **Check whether the function can get arbitrarily large.**
As $x\to\infty$, the dominant terms are:
- numerator $\sqrt{6x^2+11x-7}\approx \sqrt{6}\,|x|$,
- denominator $\sqrt{|2x-5|+4x^2-2x-13}\approx \sqrt{4x^2}=2|x|$.
So the function approaches:
$$\frac{\sqrt{6}\,|x|}{2|x|}=\frac{\sqrt{6}}{2}$$
This shows the function does not blow up to infinity for large $x$; it approaches a finite positive number.
5. **Find the exact behavior on each domain piece.**
For $x\ge \tfrac{5}{2}$, the denominator simplifies to $\sqrt{4x^2-18}$, so
$$y^2=\frac{6x^2+11x-7}{4x^2-18}$$
As $x\to\infty$, this approaches $\tfrac{6}{4}=\tfrac{3}{2}$, so
$$y\to \sqrt{\tfrac{3}{2}}=\frac{\sqrt{6}}{2}$$
For $x\le -\tfrac{7}{3}$, the function is continuous on that interval and equals $0$ at $x=-\tfrac{7}{3}$, then stays positive afterward.
6. **Conclusion for the range.**
The smallest value is $0$, and the function gets arbitrarily close to $\frac{\sqrt{6}}{2}$ from below and can take values above $0$.
So the range is:
$$[0,\infty)$$
7. **Final answer.**
The range is $[0,\infty)$.