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jeka94
4 years ago
8

What is f(x) - g(x)?

Mathematics
1 answer:
Gala2k [10]4 years ago
7 0
f(x)=9x^3+2x^2-5x+4\\\\g(x)=5x^3-7x+4\\\\f(x)-g(x)=(9x^3+2x^2-5x+4)-(5x^3-7x+4)\\\\=9x^3+2x^2-5x+4-5x^3+7x-4=9x^3-5x^3+2x^2-5x+7x+4-4\\\\=\boxed{4x^3+2x^2+2x}
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Given r(x) = 11/ (x - 42)
julsineya [31]

For a given function f(x) we define the domain restrictions as values of x that we can not use in our function. Also, for a function f(x) we define the inverse g(x) as a function such that:

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

<u>The restriction is:</u>

x ≠ 4

<u>The inverse is:</u>

y = 4 + \sqrt{\frac{11}{x} }

Here our function is:

f(x) = \frac{11}{(x - 4)^2}

We know that we can not divide by zero, so the only restriction in this function will be the one that makes the denominator equal to zero.

(x - 4)^2 = 0

x - 4 = 0

x = 4

So the only value of x that we need to remove from the domain is x = 4.

To find the inverse we try with the general form:

g(x) = a + \sqrt{\frac{b}{x} }

Evaluating this in our function we get:

g(f(x)) = a + \sqrt{\frac{b}{f(x)} }  = a + \sqrt{\frac{b*(x - 4)^2}{11 }}\\\\g(f(x)) = a + \sqrt{\frac{b}{11 }}*(x - 4)

Remember that the thing above must be equal to x, so we get:

g(f(x)) = a + \sqrt{\frac{b}{11 }}*(x - 4) = x\\\\{\frac{b}{11 }} = 1\\{\frac{b}{11 }}*4 - a = 0

From the two above equations we find:

b = 11

a = 4

Thus the inverse equation is:

y = 4 + \sqrt{\frac{11}{x} }

If you want to learn more, you can read:

brainly.com/question/10300045

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you are to give a client one tablet labeled 0.15 mg and one labeled 0.025 mg. what is the total dosage of these two tablets
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Simply add the two dosages together, (0.15+0.025) and the answer is 0.175 :)
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Does anybbody know the answer to this? #help
Gwar [14]

Answer:

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thank you

Step-by-step explanation:

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