Найти область определение функци $\newline$ $y=\sqrt{x^{2}+5x-6}$

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Find derivatives using the quotient rule II

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Q. Найти область определение функци

\newline

y=\sqrt{x^{2}+5x-6}

Understand Domain of Square Root Function: Understand the domain of a square root function. The domain of a square root function $y = \sqrt{f(x)}$ includes all values of $x$ for which the expression inside the square root, $f(x)$ , is greater than or equal to $0$ , because the square root of a negative number is not a real number.
Set Inequality for Square Root: Set the inside of the square root greater than or equal to zero. $\newline$ We need to find the values of $x$ for which $x^2 + 5x - 6$ is greater than or equal to zero. So we set up the inequality: $\newline$ $x^2 + 5x - 6 \geq 0$
Factor Quadratic Expression: Factor the quadratic expression. $\newline$ To solve the inequality, we first factor the quadratic expression $x^2 + 5x - 6$ . This factors into $(x + 6)(x - 1)$ . $\newline$ So the inequality becomes: $\newline$ $(x + 6)(x - 1) \geq 0$
Find Critical Points: Find the critical points of the inequality. $\newline$ The critical points are the values of $x$ where the expression equals zero. We set each factor equal to zero and solve for $x$ : $\newline$ $x + 6 = 0 \Rightarrow x = -6$ $\newline$ $x - 1 = 0 \Rightarrow x = 1$ $\newline$ These are the points where the expression inside the square root changes sign.
Determine Intervals to Test: Determine the intervals to test. $\newline$ We now have three intervals to test based on the critical points: $(-\infty, -6)$ , $(-6, 1)$ , and $(1, \infty)$ . We will test a value from each interval to see if the expression inside the square root is positive or negative in that interval.
Test Intervals: Test the intervals. $\newline$ For the interval $(-\infty, -6)$ , we can test $x = -7$ : $\newline$ $(-7 + 6)(-7 - 1) = (-1)(-8) = 8$ , which is positive. $\newline$ For the interval $(-6, 1)$ , we can test $x = 0$ : $\newline$ $(0 + 6)(0 - 1) = (6)(-1) = -6$ , which is negative. $\newline$ For the interval $(1, \infty)$ , we can test $x = 2$ : $\newline$ $(2 + 6)(2 - 1) = (8)(1) = 8$ , which is positive. $\newline$ So the expression inside the square root is positive in the intervals $(-\infty, -6)$ and $(1, \infty)$ .
Write Domain of Function: Write the domain of the function. $\newline$ The domain of the function $y = \sqrt{x^2 + 5x - 6}$ is the union of the intervals where the expression inside the square root is non-negative. Therefore, the domain is: $\newline$ $(-\infty, -6] \cup [1, \infty)$