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For the function below, find a) the critical numbers; b) the open intervals where the function is increasing; and c) the open intervals where it is decreasing.

f(x)=(x+7)/(x+2)

For the function below, find a) the critical numbers; b) the open intervals where the function is increasing; and c) the open intervals where it is decreasing. $\newline$ $f(x)=\frac{x+7}{x+2}$

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Find equations of tangent lines using limits

Full solution

Q. For the function below, find a) the critical numbers; b) the open intervals where the function is increasing; and c) the open intervals where it is decreasing.

\newline

f(x)=\frac{x+7}{x+2}

Find Critical Numbers: To find the critical numbers, we need to find where the derivative of $f(x)$ is $0$ or undefined.
Differentiate and Solve: Differentiate $f(x)$ with respect to $x$ to get $f'(x)$ .
$f'(x) = \frac{(x+2)(1) - (x+7)(1)}{(x+2)^2}$
$f'(x) = \frac{x+2 - x - 7}{(x+2)^2}$
$f'(x) = \frac{-5}{(x+2)^2}$
Check Derivative: The derivative $f'(x)$ is never zero because the numerator is a constant $-5$ . However, it is undefined when the denominator is zero.
Identify Critical Number: Set the denominator equal to zero and solve for $x$ . $(x+2)^2 = 0$ $x+2 = 0$ $x = -2$
Determine Increasing/Decreasing: The critical number is $x = -2$ because that's where the derivative is undefined.
Choose Test Points: Now, we need to determine where the function is increasing or decreasing. We do this by testing intervals around the critical number.
Test Point $-3$ : Choose test points in the intervals $(-\infty, -2)$ and $(-2, \infty)$ , like $x = -3$ and $x = 0$ , and plug them into $f'(x)$ .
Test Point $0$ : For $x = -3$ : $f'(-3) = (-5) / (-3+2)^2 = -5 / 1 = -5$ , which is negative.
Test Point $0$ : For $x = -3$ : $f'(-3) = (-5) / (-3+2)^2 = -5 / 1 = -5$ , which is negative.For $x = 0$ : $f'(0) = (-5) / (0+2)^2 = -5 / 4$ , which is also negative.

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