let ( f(x)=5x - 6 ) and ( g(x)=1 + x ). find the following.\n(a) ( (f + g)(x) )\n(b) ( (f - g)(x) )\n(c) (…

Explanation:

Step1: Find ((f - g)(x))

By the definition of ((f - g)(x)=f(x)-g(x)). Substitute (f(x) = 5x-6) and (g(x)=1 + x) into the formula: ((f - g)(x)=(5x - 6)-(1 + x))

Step2: Simplify the expression

[ \begin{align*} (5x - 6)-(1 + x)&=5x-6 - 1-x\ &=(5x-x)+(-6 - 1)\ &=4x-7 \end{align*} ]

Step3: Find ((f\cdot g)(x))

By the definition of ((f\cdot g)(x)=f(x)\cdot g(x)). Substitute (f(x) = 5x-6) and (g(x)=1 + x) into the formula: ((f\cdot g)(x)=(5x - 6)(1 + x))

Step4: Expand the product

[ \begin{align*} (5x - 6)(1 + x)&=5x\times1+5x\times x-6\times1-6\times x\ &=5x+5x^{2}-6 - 6x\ &=5x^{2}+(5x-6x)-6\ &=5x^{2}-x - 6 \end{align*} ]

Step5: Find (\frac{f}{g}(x))

By the definition of (\frac{f}{g}(x)=\frac{f(x)}{g(x)}), where (g(x)\neq0). Substitute (f(x) = 5x-6) and (g(x)=1 + x) into the formula: (\frac{f}{g}(x)=\frac{5x - 6}{1 + x}), with (x\neq - 1)

Step6: Find the domain of (\frac{f}{g}(x))

The domain of (\frac{f}{g}(x)) is all real numbers (x) such that the denominator (g(x)\neq0). Set (g(x)=1 + x\neq0), solve for (x): (x\neq - 1)

Answer:

(a) ((f + g)(x)=6x-1) (b) ((f - g)(x)=4x - 7) (c) ((f\cdot g)(x)=5x^{2}-x - 6) (d) (\frac{f}{g}(x)=\frac{5x - 6}{1 + x},x\neq - 1) (e) Domain of (\frac{f}{g}(x)) is ({x|x\neq - 1})