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:[tex]f(x)= \frac{9-x^2}{x^2-4}[/tex]
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 [tex]f(-4)= \frac{9-(-4)^2}{(-4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12} <0[/tex]
⇒ f (x) < 0 for x ∈ (-∞ ,-3)
b. (-3, -2)
put k = -2.5 from the given interval
We get [tex]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[/tex]
⇒ f (x) > 0 for x ∈ (-3,-2)
c. (-2, 2)
put k = 0 from the given interval
We get [tex]f(0)= \frac{9-(0)^2}{(0)^2-4} = \frac{9}{-4} = -\frac{9}{4} < 0[/tex]
⇒ f (x) < 0 for x ∈ (-2,2)
d. (2, 3)
put k =2.5 from the given interval
We get [tex]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[/tex]
⇒ f (x) > 0 for x ∈ (2,3)
e. (infinity, 3)
put k = 4 from the given interval
We get [tex]f(4)= \frac{9-(4)^2}{(4)^2-4} = \frac{9-16}{16-4} = \frac{-7}{12} <0[/tex]
⇒ f (x) < 0 for x ∈ (3,∞)
