Given $$ f(x)=x^{2}-3 x-1 $$ and $$ g(x)=x-3 $$, find $$ (f+g)(x)=\square $$ $$ (f-g)(x)=\square $$ $$ (f g)(x)=\square $$ $$ \left(\frac{f}{g} ight)(x)=\square $$ where $$ x
eq \square $$ Submit Question

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Given $$ f(x)=x^{2}-3 x-1 $$ and $$ g(x)=x-3 $$, find $$ (f+g)(x)=\square $$ $$ (f-g)(x)=\square $$ $$ (f g)(x)=\square $$ $$ \left(\frac{f}{g}ight)(x)=\square $$ where $$ x 
eq \square $$ Submit Question

UpStudy AI · Accepted Answer

Let
$$
f(x)=x^2-3x-1 \quad \text{and} \quad g(x)=x-3.
$$

1. Compute $$(f+g)(x)$$:
   $$
   (f+g)(x)=f(x)+g(x)=\left(x^2-3x-1\right)+\left(x-3\right).
   $$
   Combine like terms:
   $$
   x^2-3x-1+x-3=x^2-2x-4.
   $$
   Thus, 
   $$
   (f+g)(x)=x^2-2x-4.
   $$

2. Compute $$(f-g)(x)$$:
   $$
   (f-g)(x)=f(x)-g(x)=\left(x^2-3x-1\right)-\left(x-3\right).
   $$
   Distribute the negative sign and combine like terms:
   $$
   x^2-3x-1-x+3=x^2-4x+2.
   $$
   Thus,
   $$
   (f-g)(x)=x^2-4x+2.
   $$

3. Compute $$(fg)(x)$$:
   $$
   (fg)(x)=f(x)g(x)=\left(x^2-3x-1\right)(x-3).
   $$
   Expand using the distributive property:
   $$
   x(x^2-3x-1)-3(x^2-3x-1)=x^3-3x^2-x-3x^2+9x+3.
   $$
   Combine like terms:
   $$
   x^3-6x^2+8x+3.
   $$
   Thus,
   $$
   (fg)(x)=x^3-6x^2+8x+3.
   $$

4. Compute $$\left(\frac{f}{g}\right)(x)$$:
   $$
   \left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x^2-3x-1}{x-3},
   $$
   with the domain restriction that the denominator cannot be zero. Thus, $$x-3\neq0$$ or
   $$
   x\neq3.
   $$

Final answers:
$$
\begin{aligned}
(f+g)(x)&=x^2-2x-4,\[1mm]
(f-g)(x)&=x^2-4x+2,\[1mm]
(fg)(x)&=x^3-6x^2+8x+3,\[1mm]
\left(\frac{f}{g}\right)(x)&=\frac{x^2-3x-1}{x-3},\quad x\neq3.
\end{aligned}
$$
$$ \begin{aligned} (f+g)(x) &= x^2 - 2x - 4, \ (f-g)(x) &= x^2 - 4x + 2, \ (fg)(x) &= x^3 - 6x^2 + 8x + 3, \ \left(\frac{f}{g}\right)(x) &= \frac{x^2 - 3x - 1}{x - 3}, \quad x \neq 3. \end{aligned} $$

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