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nordsb [41]
4 years ago
10

Need help on these, thank you!

Mathematics
1 answer:
tensa zangetsu [6.8K]4 years ago
8 0

Answer:

5x+2 , 2x^2+5x-2,\frac{x^2+5x}{2-x^2}

Step-by-step explanation:

We are given f(x) and g(x)

1. (f+g)(x)

(f+g)(x) = f(x) + g(x)

           = x^2+5x+2-x^2

           = 5x+2

Domain : All real numbers as it there exists a value of (f+g)(x) f every x .

2. (f-g)(x)

(f-g)(x) = f(x)-g(x)

          = x^2+5x-2+x^2

          =2x^2+5x-2

Domain : All real numbers as it there exists a value of (f-g)(x) f every x .

Part 3 .

(\frac{f}{g})(x)\\(\frac{f}{g})(x) = \frac{f(x)}{g(x)}\\=\frac{x^2+5x}{2-x^2}

Domain : In this case we see that the function is not defined for values of x for which the denominator becomes 0 or less than zero . Hence only those values of x are defined for which

2-x^2>0

or 2>x^2

   Hence taking square roots on both sides and solving inequality we get.

-\sqrt{2}

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