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
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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