Question

Please helpppppppppppp I’m confused on greater and equal to or less and equal to sign, not sure which one to use and when to use which one - pic is problem #8 (but this is the main concern)

Answer 1

N 8

we have the function

$g(x)=\frac{1}{\sqrt[]{16-x^2}}$

Remember that

the denominator cannot be equal to zero and the discriminant must be greater than or equal to zero

that means

16-x^2 > 0

solve for x

-x^2 > -16

multiply by -1 both sides

x^2 < 16

the solution is the interval (-4,4)

therefore

In this problem

the domain is the interval (-4,4)

Explanation

the domain is (-4,4)

If you put in the given function x=4

$g(x)=\frac{1}{\sqrt[]{16-4^2}}=\frac{1}{0}$

undefined

If you put x=5

$g(x)=\frac{1}{\sqrt[]{16-5^2}}=\frac{1}{\sqrt[]{-9}}$

is not a real number

For that reason, the domain is the interval (-4,4)

x^2 < 16

square root both sides

we have

(+/-)x < 4 -----> two inequalities

First

+x <4

second

-x <4 -----> x >-4

the solution is the compound inequality

x <4 and x> -4

we have the problem

$f(x)=\sqrt[]{16-x^2}$

In this case

the radicand must be greater than or equal to zero

so

$16-x^2\ge0$

in this case

the domain is the interval [-4,4]

the value of x=-4 and x=4 are included in the domain because the function can be equal to zero

In the previous problem, the values of x=-4 and x=4 are not included, because the function cannot be equal to zero