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vfiekz [6]
3 years ago
5

P(x) 3x^2 + 5x^2 + 4x-7 andq (x)=6x^3+ 2x^2-9x-5Find P(x)+ q (x) and P[x]-q(x)​

Mathematics
1 answer:
nikdorinn [45]3 years ago
6 0

Answer:

P(x) + q(x)

= (3x² + 5x² + 4x - 7) + (6x³ + 2x² - 9x - 5)

= 3x² + 5x² + 4x - 7 + 6x³ + 2x² - 9x - 5

= 6x³ + 10x² - 5x - 12

P(x) - q(x)

= (3x² + 5x² + 4x - 7) - (6x³ + 2x² - 9x - 5)

= 3x² + 5x² + 4x - 7 - 6x³ - 2x² + 9x + 5

= -6x³ + 6x² + 13x - 2

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The domain of startfraction f over g endfraction f g consists of numbers x for which g left parenthesis x right parenthesisg(x)
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The domain of f/g consists of numbers x for which g(x) cannot equal 0 that are in the domains of both f and g.

 

Let’s take this equation as an example:

If f(x) = 3x - 5 and g(x) = square root of x-5, what is the domain of (f/g)x. 


For x to be in the domain of (f/g)(x), it must be in the domain of f and in the domain of g since (f/g)(x) = f(x)/g(x). We also need to ensure that g(x) is not zero since f(x) is divided by g(x). Therefore, there are 3 conditions.

 

x must be in the domain of f: f(x) = 3x -5 are in the domain of x and all real numbers x.

x must be in the domain of g: g(x) = √(x - 5) so x - 5 ≥ 0 so x ≥ 5.

g(x) can not be 0: g(x) = √(x - 5) and √(x - 5) = 0 gives x = 5 so x ≠ 5.

 

Hence to x x ≥ 5 and x ≠ 5 so the domain of (f/g)(x) is all x satisfying x > 5.

 

Thus, satisfying <span>satisfy all three conditions, x x ≥ 5 and x ≠ 5 so the domain of (f/g)(x) is all x satisfying x > 5.</span>

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