I need to check my answer for these two

Question

Also i get confuse with greater and equal to and lesser and equal to, when is when and how do i know when to use it?

Answer 1

3) The function

Simplify the rational function.

Factorize the denominator.

x² - 2x - 3 

= x² - 3x + x - 3 

= x(x - 3) + 1 (x - 3)

= (x - 3) (x + 1)

Vertical asymptotes can be found by making denominator = 0.

 x + 1 = 0

x = - 1

Vertical asymptote is x = - 1.

Answer 2

4) The functions are f(x) = 2x² + 3x - 1 and g(x) = √(4x + 1)

f(g(x))

Substitute the expression for functioning g (in this case √(4x + 1)) for g(x) in the composition.

= f[√(4x + 1)]

Substitute the expression for functioning f (in this case 2x² + 3x - 1) for f(x) in the composition.

= 2[√(4x + 1)]² + 3[√(4x + 1)] - 1

= 2(4x + 1) + 3[√(4x + 1)] - 1

f(g(x)) = 8x + 2 + 3[√(4x + 1)]-1

f(g(2)) = 8(2) + 1 + 3[√(4(2) + 1)]

= 16 + 1 + 3 (√9)

= 16 + 1 + 9

= 26

f(g(2)) = 26.

Answer 3

5) The radical function f(x) = √(16 - x²)

Index of radical function = 2

Since the index is even, set the expression inside the radical greater than or equal to 0.

16 - x² ≥ 0

Now solve the above inequality.

16 - x² ≥ 0

Multiply each side of inequality by negative 1 and flip the symbol.

x² - 16 ≤  0

(x - 4)(x + 4) ≤  0

Now, there are two ways this product could be less than zero.One factor must be negative and one must be positive.

First situation: x - 4 < 0 and x + 4 > 0

x < 4 and x > - 4

Second situation: x - 4 > 0 and x + 4 < 0

x > 4 and x < - 4

There are NO values for which this situation is true.

Note that the inequality contains a “≤ ” symbol, We include it into set of solutions at x = 4 and - 4.

At x = 4

(x - 4)(x + 4) ≤  0

(4 - 4)(4 + 4) ≤  0

0 ≤  0

At x = - 4

(x - 4)(x + 4) ≤  0

( - 4 - 4)(- 4 + 4) ≤  0

0 ≤  0

The above statements are true.

Solution of the inequality 16 - x² ≥ 0 is - 4 ≤ x ≤ 4.

Domain of the function is [- 4, 4].

Comments

Why do you flip signs? How you determine that index radical, what does that mean?

Solution of the inequality 16 - x² ≥ 0 is - 4 ≤ x ≤ 4.

 

what that mean?