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Chapter 1: Problem 45

Find \(f+g, f-g,\) fg, and \(\frac{f}{8} .\) Determine the domain for each function. $$f(x)=\frac{9 x}{x-4}, g(x)=\frac{7}{x+8}$$

Short Answer

Expert verified
\(f + g = \frac{9x^2+79x-28}{x^2+4x-32}\), \(f - g = \frac{9x^2+65x+28}{x^2+4x-32}\), \(fg = \frac{63x}{x^2 + 4x - 32}\), \(\frac{f}{g} = \frac{9x(x + 8)}{7(x - 4)}\). The domains for all the operations are \(-\infty < x < -8, -8 < x < 4, 4 < x < \infty\).

Step by step solution

01

Calculate \(f + g\)

The function \(f + g\) is defined as \(f(x) + g(x)\). So we substitute \(f(x)\) and \(g(x)\) given in the problem into this equation. Doing this, we get: \(f + g = \frac{9x}{x-4} + \frac{7}{x+8}\). To simplify this expression, we find a common denominator and add the numerators, resulting in: \(f + g = \frac{9x(x+8) + 7(x-4)}{(x-4)(x+8)}\) = \(\frac{9x^2+72x+7x-28}{x^2+4x-32} = \frac{9x^2+79x-28}{x^2+4x-32}\)
02

Determine the Domain of \(f + g\)

The domain of the function \(f+g\) is all real numbers except for the values of x which make the denominator equal to zero. So, set \(x^2+4x-32 = 0\) and solve for \(x\). The solutions are \(x = -8\) and \(x = 4\). Therefore, the domain of \(f + g\) is \(-\infty < x < -8, -8 < x < 4, 4 < x < \infty\).
03

Calculate \(f - g\)

The function \(f - g\) is defined as \(f(x) - g(x)\). So we substitute \(f(x)\) and \(g(x)\) given in the problem into this equation. Doing this, we get: \(f - g = \frac{9x}{x-4} - \frac{7}{x+8}\). To simplify this expression, we find a common denominator and subtract the numerators, resulting in: \(f - g = \frac{9x(x+8) - 7(x-4)}{(x-4)(x+8)}\) = \(\frac{9x^2+72x-7x+28}{x^2+4x-32} = \frac{9x^2+65x+28}{x^2+4x-32}\)
04

Determine the Domain of \(f - g\)

The domain of the function \(f-g\) is all real numbers except for the values of x which make the denominator equal to zero. The solutions are \(x = -8\) and \(x = 4\). Therefore, the domain of \(f - g\) is \(-\infty < x < -8, -8 < x < 4, 4 < x < \infty\).
05

Calculate fg

The function fg is defined as \(f(x)g(x)\). So, we substitute in the functions \(f(x)\) and \(g(x)\) and we get: \(fg = \left(\frac{9x}{x-4}\right)\left(\frac{7}{x+8}\right)\) = \(\frac{63x}{(x-4)(x+8)}\) = \(\frac{63x}{x^2 + 4x - 32}\)
06

Determine the Domain of fg

The domain of the function \(fg\) is all real values except for the values of \(x\) that make the denominator equal to zero. The solutions are \(x = -8\) and \(x = 4\). Therefore, the domain of \(fg\) is \(-\infty < x < -8, -8 < x < 4, 4 < x < \infty\).
07

Calculate \(\frac{f}{g}\)

The function \(\frac{f}{g}\) is defined as \(\frac{f(x)}{g(x)}\). So, we substitute \(f(x)\) and \(g(x)\) and get: \(\frac{f}{g} = \frac{\frac{9x}{x-4}}{\frac{7}{x+8}}\). Simplifying this expression, we multiply the first fraction by the reciprocal of the second and obtain: \(\frac{f}{g} = \frac{9x(x + 8)}{7(x - 4)}\)
08

Determine the Domain of \(\frac{f}{g}\)

The domain of the function \(\frac{f}{g}\) is all real values except for values of \(x\) that make the denominator equal to zero and those making the \(g\) equal to zero. Solve \(x - 4 = 0\) and \(x + 8 = 0\) to find the solutions are \(x = 4\) and \(x = -8\). Therefore, the domain of \(\frac{f}{g}\) is \(-\infty < x < -8, -8 < x < 4, 4 < x < \infty\).

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