# Domain of One Variable Function – A function with two branches – Exercise 5755

Exercise

Determine the domain of the function:

$$f(x) = \begin{cases} \frac{\sqrt{2-x}-\sqrt{4-3x}}{x^2-3x+2}, &\quad x<1\\ \frac{2x^2-6x+4}{x^2-1}, &\quad x>1\\ \end{cases}$$

$$x<1\text{ or }x>1$$

Solution

We find the domain of the function. Since there is a denominator, we require that the expression in the denominator be different from zero:

$$x^2-3x+2\neq 0$$

$$x^2-1\neq 0$$

Also, there are square roots, so the expressions inside the roots must be non-negative:

$$2-x\geq 0$$

$$4-3x\geq 0$$

We solve all inequalities and intersect their results (“and”).

Solve the first inequality:

$$x^2-3x+2\neq 0$$

It is a quadratic equation. Its coefficients are:

$$a=1, b=-3, c=2$$

We find the solutions (= zeros = roots) of the quadratic equation using the quadratic formula. Putting the coefficients in the formula gives us

$$x_{1,2}=\frac{3\pm \sqrt{{(-3)}^2-4\cdot 1\cdot 2}}{2\cdot 1}=$$

$$=\frac{3\pm \sqrt{1}}{2}=$$

$$=\frac{3\pm 1}{2}$$

Hence, we get the solutions:

$$x_1=\frac{3+ 1}{2}=2$$

$$x_2=\frac{3-1}{2}=1$$

Since we are interested in the section where the parabola is different from zero, the answer is

$$x\neq 1,2$$

Now, we intersect the result with the function domain (in the exercise):

$$x<1$$

And we will get that the solution of the first inequality is

$$x<1$$

Solve the second inequality:

$$x^2-1\neq 0$$

It is again a square inequality. Its roots are:

$$x= \pm 1$$

Since we are interested in the section where the parabola is different from zero, the answer is

$$x\neq \pm 1$$

Again, we intersect the result with the function domain (in the exercise):

$$x>1$$

And we will get that the solution of the second inequality is

$$x>1$$

Solve the third inequality:

$$2-x\geq 0$$

$$x\leq 2$$

We intersect the result with the function domain (in the exercise):

$$x<1$$

And we will get that the solution of the third inequality is

$$x<1$$

Solve the fourth inequality:

$$4-3x\geq 0$$

$$3x\leq 4$$

$$x\leq \frac{4}{3}$$

We intersect the result with the function domain (in the exercise):

$$x<1$$

And we will get that the solution of the fourth inequality is

$$x<1$$

Finally, we intersect all the results we got, and the final answer is

$$x<1\text{ or }x>1$$

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