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il63 [147K]
3 years ago
15

You must find all intervals where f (x) is positive. F(x)=-3x^2-6x+24

Mathematics
1 answer:
Lorico [155]3 years ago
6 0

Answer:

he positive regions of a function are those intervals where the function is above the x-axis. It is where the y-values are positive (not zero). The negative regions of a function are those intervals where the function is below the x-axis. It is where the y-values are negative

Step-by-step explanation:

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<h2>D. \frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\ or StartFraction x squared + 3 x minus 12 Over (x + 3) (x minus 5) (x + 7) EndFraction</h2>

Step-by-step explanation:

Given the expression \frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }, the dfference is expressed as follows;

Step1: First we need to factorize the denominator of each function.

\frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }\\= \frac{x}{x^{2}-5x+3x-15 } - \frac{4}{x^{2} + 7x-5x - 35 }\\= \frac{x}{x(x-5)+3(x-5) } - \frac{4}{x( x+ 7)x-5(x +7) }\\= \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\\\

Step 2: We will find the LCM of the resulting expression

=  \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\= \frac{x(x+7)-4(x+3)}{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+7x-4x-12 }{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\

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