Questions: Let f(x) = 2x / (2x + 5) and g(x) = (13x + 15) / (2x^2 + 3x - 5). Simplify f(x) + g(x) : The values of x that need to be excluded from the domain of the sum function are

Solution Steps

To simplify \( f(x) + g(x) \), we need to find a common denominator and combine the fractions. Then, we simplify the resulting expression. For the domain, we need to identify the values of \( x \) that make the denominators zero.

Given: \[ f(x) = \frac{2x}{2x + 5} \] \[ g(x) = \frac{13x + 15}{2x^2 + 3x - 5} \]

First, we need to find a common denominator for the two fractions. The denominators are \(2x + 5\) and \(2x^2 + 3x - 5\).

Factorize the denominator of \(g(x)\): \[ 2x^2 + 3x - 5 = (2x - 1)(x + 5) \]

So, the common denominator is: \[ (2x + 5)(2x - 1)(x + 5) \]

Rewrite \(f(x)\) and \(g(x)\) with the common denominator: \[ f(x) = \frac{2x}{2x + 5} \cdot \frac{(2x - 1)(x + 5)}{(2x - 1)(x + 5)} = \frac{2x(2x - 1)(x + 5)}{(2x + 5)(2x - 1)(x + 5)} \] \[ g(x) = \frac{13x + 15}{(2x - 1)(x + 5)} \cdot \frac{2x + 5}{2x + 5} = \frac{(13x + 15)(2x + 5)}{(2x + 5)(2x - 1)(x + 5)} \]

Now, add the two fractions: \[ f(x) + g(x) = \frac{2x(2x - 1)(x + 5) + (13x + 15)(2x + 5)}{(2x + 5)(2x - 1)(x + 5)} \]

Simplify the numerator: \[ 2x(2x - 1)(x + 5) = 2x(2x^2 + 9x - 5) = 4x^3 + 18x^2 - 10x \] \[ (13x + 15)(2x + 5) = 26x^2 + 65x + 30x + 75 = 26x^2 + 95x + 75 \]

Combine the numerators: \[ 4x^3 + 18x^2 - 10x + 26x^2 + 95x + 75 = 4x^3 + 44x^2 + 85x + 75 \]

Thus: \[ f(x) + g(x) = \frac{4x^3 + 44x^2 + 85x + 75}{(2x + 5)(2x - 1)(x + 5)} \]

The values of \( x \) that make the denominator zero must be excluded. Set each factor of the denominator to zero and solve for \( x \):

\[ 2x + 5 = 0 \implies x = -\frac{5}{2} \] \[ 2x - 1 = 0 \implies x = \frac{1}{2} \] \[ x + 5 = 0 \implies x = -5 \]

Final Answer