Question

Consider the function f(x)=9-x^2/x^2-4 For which intervals is f(x) positive? Check ALL that apply.

Answer 1

Answer:

a. f (x) < 0 for x ∈ (-∞ ,-3)

b. f (x) > 0 for x ∈ (-3,-2)

c. f (x) < 0 for x ∈ (-2,2)

d. f (x) > 0 for x ∈ (2,3)

e.f (x) < 0 for x ∈ (3,∞)

Step-by-step explanation:

Here, the given function is:$f(x)= \frac{9-x^2}{x^2-4}$

Now, to check for the sign of f(x) at x = k, put the value of x from the given interval.

We get:

a. (-infinity, -3)

put k = -4 from the given interval

We get $f(-4)= \frac{9-(-4)^2}{(-4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12}$

⇒ f (x) < 0 for x ∈ (-∞ ,-3)

b. (-3, -2)

put k = -2.5 from the given interval

We get $f(-2.5)= \frac{9-(-2.5)^2}{(-2.5)^2-4} = \frac{9-6.25}{6.25-4} = \frac{2.75}{2.25} > 0$

⇒ f (x) > 0 for x ∈ (-3,-2)

c. (-2, 2)

put k = 0 from the given interval

We get $f(0)= \frac{9-(0)^2}{(0)^2-4} = \frac{9}{-4} = -\frac{9}{4} < 0$

⇒ f (x) < 0 for x ∈ (-2,2)

d. (2, 3)

put k =2.5 from the given interval

We get $f(2.5)= \frac{9-(2.5)^2}{(2.5)^2-4} = \frac{9-6.25}{6.25-4} = \frac{2.75}{2.25} > 0$

⇒ f (x) > 0 for x ∈ (2,3)

e. (infinity, 3)

put k = 4 from the given interval

We get $f(4)= \frac{9-(4)^2}{(4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12}$

⇒ f (x) < 0 for x ∈ (3,∞)