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The graph of
f(x)=(4)/(x^(2)-2x-3) is shown. For which values of
x is
f(x) decreasing?

The graph of $f(x)=\frac{4}{x^{2}-2 x-3}$ is shown. For which values of $x$ is $f(x)$ decreasing?

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Find the vertex of the transformed function

Full solution

Q. The graph of

f(x)=\frac{4}{x^{2}-2 x-3}

is shown. For which values of

x

is

f(x)

decreasing?

Analyze function and domain: Step $1$ : Analyze the function and its domain. $\newline$ $f(x) = \frac{4}{x^2 - 2x - 3}$ $\newline$ Factorize the denominator. $\newline$ $x^2 - 2x - 3 = (x - 3)(x + 1)$ $\newline$ The function is undefined at $x = 3$ and $x = -1$ .
Determine intervals for behavior: Step $2$ : Determine the intervals to test for increasing or decreasing behavior. The critical points are $x = 3$ and $x = -1$ , splitting the real line into three intervals: $(-\infty, -1)$ , $(-1, 3)$ , and $(3, \infty)$ .
Choose test points for derivative: Step $3$ : Choose test points from each interval and plug them into the derivative $f'(x)$ . $f'(x) = -\left[\frac{8x - 8}{(x^2 - 2x - 3)^2}\right]$ Test points: $x = -2, 0,$ and $4$ . $f'(-2) = -\left[\frac{8(-2) - 8}{((-2)^2 - 2(-2) - 3)^2}\right] = -\left[\frac{8(-2) - 8}{(4 + 4 - 3)^2}\right] = -\left[\frac{8(-2) - 8}{(5)^2}\right] = -\left[\frac{(-24)}{25}\right]$ $f'(0) = -\left[\frac{8(0) - 8}{(0^2 - 2(0) - 3)^2}\right] = -\left[\frac{(-8)}{9}\right]$ $f'(4) = -\left[\frac{8(4) - 8}{(4^2 - 2(4) - 3)^2}\right] = -\left[\frac{(24)}{49}\right]$

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