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Leto [7]
3 years ago
14

Use the given functions f and g to find f + g, f − g, fg, and f g . State the domain of each. (Enter your answer for the domain

in interval notation.) f(x) = 3x + 6, g(x) = x + 2.
Find the domain of each problem.
f + g = Domain=
f-g= Domain=
(f)(g)= Domain=
f/g=Domain=
2.) Find (g ○ f)(x) and (f ○ g)(x) for the given functions f and g.
f(x) = 3/(x+5), g(x) = 3x − 6
(g ○ f)(x) =
(f ○ g)(x) =
3.) Use the method of completing the square to find the standard form of the quadratic function.
f(x) = x2 − 8x + 2
y =
4.) Use the vertex formula to determine the vertex of the graph of the function.
f(x) = x2 − 14x Write the function in standard form.
f(x) =
5.) Use the vertex formula to determine the vertex of the graph of the function.
f(x) = 3x2 − 10x + 1 Write the function in standard form.
f(x) =

Mathematics
1 answer:
xeze [42]3 years ago
4 0

Answer: The answers are stated below

Step-by-step explanation: Attached below is the explaination of the solution.

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Exponential Equation WITHOUT CALCULATOR
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Multiply both sides of the first equation by 2^x:

2^x-2^{-x}=4\implies 2^{2x}-1=4\cdot2^x\implies 2^{2x}-4\cdot2^x-1=0

This is quadratic in 2^x; to make this clear, substitute y=2^x. Then

y^2-4y-1=0\implies y=2\pm\sqrt5

One of these solutions for y is negative. But, if x is real, then y=2^x is always supposed to be positive, so we can throw out the negative root, leaving y=2+\sqrt5.

We actually don't have to solve for x exactly. We can just rewrite the next two equations in terms of y.


2^{2x}+2^{-2x}=(2^x)^2+(2^{-x})^2=y^2+\dfrac1{y^2}

2^{3x}-2^{-3x}=y^3-\dfrac1{y^3}


Since y=2+\sqrt5, we get

(2+\sqrt5)^2+\dfrac1{(2+\sqrt5)^2}=18

(2+\sqrt5)^3-\dfrac1{(2+\sqrt5)^3}=76
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