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

Exercise

Determine the domain of the function:

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

Final Answer

Show final answer

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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