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Math  /  Algebra

QuestionFind f+g,f−g,fg\mathrm{f}+\mathrm{g}, \mathrm{f}-\mathrm{g}, \mathrm{fg}, and fg\frac{\mathrm{f}}{\mathrm{g}}. Determine the domain for each function. f(x)=x;g(x)=x−20f(x)=\sqrt{x} ; g(x)=x-20 (f+g)(x)=(f+g)(x)= □\square (Simplify your aniswer.)

Studdy Solution

STEP 1

1. We are given two functions: f(x)=x f(x) = \sqrt{x} and g(x)=x−20 g(x) = x - 20 .
2. We need to find the sum, difference, product, and quotient of these functions.
3. The domain of each resulting function will be determined by the restrictions of both f(x) f(x) and g(x) g(x) .

STEP 2

1. Calculate (f+g)(x) (f+g)(x) and determine its domain.
2. Calculate (f−g)(x) (f-g)(x) and determine its domain.
3. Calculate (fg)(x) (fg)(x) and determine its domain.
4. Calculate (fg)(x) \left(\frac{f}{g}\right)(x) and determine its domain.

STEP 3

Calculate (f+g)(x) (f+g)(x) :
(f+g)(x)=f(x)+g(x)=x+(x−20)=x+x−20 (f+g)(x) = f(x) + g(x) = \sqrt{x} + (x - 20) = \sqrt{x} + x - 20
Determine the domain of (f+g)(x) (f+g)(x) :
The domain of f(x)=x f(x) = \sqrt{x} is x≥0 x \geq 0 because the square root function is only defined for non-negative numbers. The domain of g(x)=x−20 g(x) = x - 20 is all real numbers. Therefore, the domain of (f+g)(x) (f+g)(x) is x≥0 x \geq 0 .

STEP 4

Calculate (f−g)(x) (f-g)(x) :
(f−g)(x)=f(x)−g(x)=x−(x−20)=x−x+20 (f-g)(x) = f(x) - g(x) = \sqrt{x} - (x - 20) = \sqrt{x} - x + 20
Determine the domain of (f−g)(x) (f-g)(x) :
Similar to (f+g)(x) (f+g)(x) , the domain of (f−g)(x) (f-g)(x) is restricted by f(x)=x f(x) = \sqrt{x} , so the domain is x≥0 x \geq 0 .

STEP 5

Calculate (fg)(x) (fg)(x) :
(fg)(x)=f(x)⋅g(x)=x⋅(x−20)=x(x−20) (fg)(x) = f(x) \cdot g(x) = \sqrt{x} \cdot (x - 20) = \sqrt{x}(x - 20)
Determine the domain of (fg)(x) (fg)(x) :
The domain is again restricted by f(x)=x f(x) = \sqrt{x} , so the domain is x≥0 x \geq 0 .

STEP 6

Calculate (fg)(x) \left(\frac{f}{g}\right)(x) :
(fg)(x)=f(x)g(x)=xx−20 \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x}}{x - 20}
Determine the domain of (fg)(x) \left(\frac{f}{g}\right)(x) :
1. The domain of f(x)=x f(x) = \sqrt{x} is x≥0 x \geq 0 .
2. The function g(x)=x−20 g(x) = x - 20 cannot be zero, so x−20≠0 x - 20 \neq 0 , which implies x≠20 x \neq 20 .
Therefore, the domain of (fg)(x) \left(\frac{f}{g}\right)(x) is x≥0 x \geq 0 and x≠20 x \neq 20 , or in interval notation: [0,20)∪(20,∞) [0, 20) \cup (20, \infty) .
The solutions are: - (f+g)(x)=x+x−20 (f+g)(x) = \sqrt{x} + x - 20 with domain x≥0 x \geq 0 . - (f−g)(x)=x−x+20 (f-g)(x) = \sqrt{x} - x + 20 with domain x≥0 x \geq 0 . - (fg)(x)=x(x−20) (fg)(x) = \sqrt{x}(x - 20) with domain x≥0 x \geq 0 . - (fg)(x)=xx−20 \left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x - 20} with domain [0,20)∪(20,∞) [0, 20) \cup (20, \infty) .

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