Question

Consider the function . For which intervals is f(x) positive? Check all that apply. (–∞, –3) (–3, –2) (–2, 2) (2, 3) (∞, 3)

Answer 1

Note: As you missed to add the function, so after a little research I was able to find the function. It will anyways help you in terms of clearing your concept about the context.

Answer:

For the intervals (–3, –2) and (2, 3), the function $\:f\left(x\right)=\frac{9-x^2}{x^2-4}$ will be positive. It is also clear from the graph attached below.

Step-by-step explanation:

Considering the function

$\:f\left(x\right)=\frac{9-x^2}{x^2-4}$

The intervals  (∞, 3) and (–2, 2) will make the function negative. As 1 is between the interval (∞, 3) as well as between the interval (–2, 2).

For example, putting x = 1 in the function

$\:f\left(x\right)=\frac{9-x^2}{x^2-4}$

$f(x)=\frac{9-1^2}{1^2-4}$

$\mathrm{Apply\:rule}\:1^a=1$

$1^2=1$

$f(x)=\frac{9-1}{1-4}$

$\mathrm{Subtract\:the\:numbers:}\:9-1=8$

$f(x)=\frac{8}{1-4}$

$\mathrm{Subtract\:the\:numbers:}\:1-4=-3$

$f(x)=\frac{8}{-3}$

$f(x)=-\frac{8}{3}$

Thus, the values between the intervals  (∞, 3) and (–2, 2) will make the function negative. It is also clear from the graph attached below.

Similarly, the values between the interval (–∞, –3) will also make $\:f\left(x\right)=\frac{9-x^2}{x^2-4}$ negative.

For example, putting x = -5 in the function

$\:f\left(x\right)=\frac{9-x^2}{x^2-4}$

$f(x)=\frac{9-\left(-5\right)^2}{\left(-5\right)^2-4}$

$f(x)=\frac{-16}{\left(-5\right)^2-4}$

$f(x)=\frac{-16}{21}$

$f(x)=-\frac{16}{21}$

Thus,  the values between the interval (–∞, –3) will also make $\:f\left(x\right)=\frac{9-x^2}{x^2-4}$ negative. It is also clear from the graph attached below.

The values between the the interval (–3, –2) will make the function positive. For example, putting x = -2.5 in the function,

$\:f\left(x\right)=\frac{9-x^2}{x^2-4}$

$f(x)=\frac{9-\left(-2.5\right)^2}{\left(-2.5\right)^2-4}$

$f(x)=\frac{2.75}{\left(-2.5\right)^2-4}$

$f(x)=\frac{2.75}{2.25}$

$f(x)=1.22$

Thus, the values between the the interval (–3, –2) will make the function positive. It is also clear from the graph attached below.

Similarly, the values between the interval (2, 3) will also make $\:f\left(x\right)=\frac{9-x^2}{x^2-4}$ positive. For example, putting x = 2.5 in the function,

$\:f\left(x\right)=\frac{9-x^2}{x^2-4}$

$f(x)=\frac{9-\left(2.5\right)^2}{\left(2.5\right)^2-4}$

$f(x)=\frac{2.75}{2.5^2-4}$

$f(x)=\frac{2.75}{2.25}$

$f(x)=1.22$

Thus, the values between the the interval (2, 3) will make the function positive. It is also clear from the graph attached below.

Therefore, we can CONCLUDE that for the intervals (–3, –2) and (2, 3), the function $\:f\left(x\right)=\frac{9-x^2}{x^2-4}$ will be positive. It is also clear from the graph attached below.

Keywords: equation , interval, graph

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Answer 2

Answer: 2 and 4

(–3, –2)

(2, 3)

Step-by-step explanation: