Question

1. find the domain of the function.

\\(\frac{x}{\sqrt{x - 5}}\\)

1. find the value for the function.

find \\(f(x + 1)\\) when \\(f(x) = \frac{x^2 - 3}{x + 5}\\).

1. find the domain of the function.

\\(g(x) = \frac{2x}{x^2 - 9}\\)

Explanation:

Question 5

Step 1: Analyze the square root

For the square root \( \sqrt{x - 5} \), the expression inside the square root (the radicand) must be non - negative. So we have the inequality \( x - 5\geq0 \), which gives \( x\geq5 \). But we also have a square root in the denominator.

Step 2: Analyze the denominator

Since the square root is in the denominator, \( \sqrt{x - 5}
eq0 \). If \( \sqrt{x - 5}=0 \), then \( x - 5 = 0\) or \( x = 5 \). So we need to exclude \( x = 5 \) from the values we got from the square root analysis. Combining these two conditions, we get \( x-5>0 \) (because the radicand must be positive, not just non - negative, to avoid division by zero). Solving \( x - 5>0 \) gives \( x>5 \).

Question 6

Step 1: Substitute \( x+1 \) into the function

Given \( f(x)=\frac{x^{2}-3}{x + 5} \), to find \( f(x + 1) \), we replace every occurrence of \( x \) in the function \( f(x) \) with \( x + 1 \). So we substitute \( x=x + 1 \) into the numerator and the denominator.
The numerator becomes \( (x + 1)^{2}-3 \), and the denominator becomes \( (x + 1)+5 \).

Step 2: Simplify the numerator

Expand \( (x + 1)^{2} \) using the formula \( (a + b)^{2}=a^{2}+2ab + b^{2} \), where \( a=x \) and \( b = 1 \). So \( (x + 1)^{2}=x^{2}+2x + 1 \). Then \( (x + 1)^{2}-3=x^{2}+2x+1 - 3=x^{2}+2x - 2 \).

Step 3: Simplify the denominator

Simplify \( (x + 1)+5=x + 6 \). So \( f(x + 1)=\frac{x^{2}+2x - 2}{x + 6} \), and we must note that the denominator \( x+6
eq0 \), so \( x
eq - 6 \).

Question 7

Step 1: Analyze the denominator

For the function \( g(x)=\frac{2x}{x^{2}-9} \), the denominator \( x^{2}-9
eq0 \).

Step 2: Solve the equation for the denominator

We know that \( x^{2}-9=(x - 3)(x + 3) \) (using the difference of squares formula \( a^{2}-b^{2}=(a - b)(a + b) \), where \( a = x \) and \( b = 3 \)). Set \( (x - 3)(x + 3)=0 \). Using the zero - product property, if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \). So \( x-3=0 \) gives \( x = 3 \) and \( x + 3=0 \) gives \( x=-3 \). So the values \( x = 3 \) and \( x=-3 \) make the denominator zero, and thus must be excluded from the domain.

Answer:

The domain is \( x > 5 \) (or in interval notation, \( (5, \infty) \))