Answer :
To determine where the function [tex]\( f(x) = \sqrt{x^2 - 4x + 7} \)[/tex] is continuous, we need to analyze the expression inside the square root. For the square root function to be real and defined, the expression under the square root must be non-negative ([tex]\(\geq 0\)[/tex]).
The expression inside the square root is:
[tex]\[ x^2 - 4x + 7 \][/tex]
We need to solve the inequality:
[tex]\[ x^2 - 4x + 7 \geq 0 \][/tex]
To understand this inequality, we start by finding the roots of the corresponding equation:
[tex]\[ x^2 - 4x + 7 = 0 \][/tex]
We use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 7 \)[/tex].
Calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 7 = 16 - 28 = -12 \][/tex]
Since the discriminant is negative ([tex]\(\Delta < 0\)[/tex]), the quadratic equation has no real roots. The roots are complex:
[tex]\[ x = \frac{4 \pm \sqrt{-12}}{2 \cdot 1} = \frac{4 \pm 2i\sqrt{3}}{2} = 2 \pm i\sqrt{3} \][/tex]
These complex roots indicate that the quadratic expression [tex]\( x^2 - 4x + 7 \)[/tex] does not intersect the x-axis and, since the coefficient of [tex]\( x^2 \)[/tex] is positive (the parabola opens upwards), the expression is always positive for all real numbers [tex]\( x \)[/tex].
Therefore, the expression inside the square root is always positive, implying that:
[tex]\[ x^2 - 4x + 7 > 0 \ \text{for all real} \ x \][/tex]
Since [tex]\( x^2 - 4x + 7 \)[/tex] is never zero and always positive, the function [tex]\( f(x) = \sqrt{x^2 - 4x + 7} \)[/tex] is defined and continuous for all real numbers.
Thus, the interval of continuity for [tex]\( f(x) \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
In summary:
The discriminant of the quadratic is negative, leading to complex roots.
The quadratic expression is always positive for all real [tex]\( x \)[/tex].
* Consequently, [tex]\( f(x) = \sqrt{x^2 - 4x + 7} \)[/tex] is continuous over the entire set of real numbers.
The interval(s) where [tex]\( f(x) \)[/tex] is continuous is:
[tex]\[ (-\infty, \infty) \][/tex]
The expression inside the square root is:
[tex]\[ x^2 - 4x + 7 \][/tex]
We need to solve the inequality:
[tex]\[ x^2 - 4x + 7 \geq 0 \][/tex]
To understand this inequality, we start by finding the roots of the corresponding equation:
[tex]\[ x^2 - 4x + 7 = 0 \][/tex]
We use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 7 \)[/tex].
Calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 7 = 16 - 28 = -12 \][/tex]
Since the discriminant is negative ([tex]\(\Delta < 0\)[/tex]), the quadratic equation has no real roots. The roots are complex:
[tex]\[ x = \frac{4 \pm \sqrt{-12}}{2 \cdot 1} = \frac{4 \pm 2i\sqrt{3}}{2} = 2 \pm i\sqrt{3} \][/tex]
These complex roots indicate that the quadratic expression [tex]\( x^2 - 4x + 7 \)[/tex] does not intersect the x-axis and, since the coefficient of [tex]\( x^2 \)[/tex] is positive (the parabola opens upwards), the expression is always positive for all real numbers [tex]\( x \)[/tex].
Therefore, the expression inside the square root is always positive, implying that:
[tex]\[ x^2 - 4x + 7 > 0 \ \text{for all real} \ x \][/tex]
Since [tex]\( x^2 - 4x + 7 \)[/tex] is never zero and always positive, the function [tex]\( f(x) = \sqrt{x^2 - 4x + 7} \)[/tex] is defined and continuous for all real numbers.
Thus, the interval of continuity for [tex]\( f(x) \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
In summary:
The discriminant of the quadratic is negative, leading to complex roots.
The quadratic expression is always positive for all real [tex]\( x \)[/tex].
* Consequently, [tex]\( f(x) = \sqrt{x^2 - 4x + 7} \)[/tex] is continuous over the entire set of real numbers.
The interval(s) where [tex]\( f(x) \)[/tex] is continuous is:
[tex]\[ (-\infty, \infty) \][/tex]