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Natalija [7]
3 years ago
6

Which statement could be used to explain why the function h(x) = x3 has an inverse relation that is also a function?

Mathematics
1 answer:
eimsori [14]3 years ago
3 0

Answer:

h(x) = x^3

A function has an inverse that is also a function is it is a one-to-one function, a function is one-to-one if each value in the domain is linked to only one value in the range, and if each value in the range is linked to only one value in the domain.

Then, a function that is monotonous growing is always one to one, and a function is monotonous growing if the derivative is positive for all the values of x

The derivative of x^3 is:

h'(x) = 3*x^2

and as you know, x^2 is always equal or greater than zero, so h(x) = x^3 is monotonous growing, then it has a inverse that is also a function.

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Answer: A'=(1, 3); B'=(-3, 4);C'=(3, 0); D'=(-2, 5)

You can check the PNG attached as well.

Step-by-step explanation:

You need to represent the symmetry of every given points respet to the line

y = 2

In that case, the line beeing paralell to the x- axis, x- value of the symmetry is the same of the given point and y = 2 is the middle between both points.

Point A(1, 1)

x_{A} = 1\\ x_{A'} = 1 \\\\\frac{y_{A} +y_{A'} }{2} =2\\y_{A'} = 4 - y_{A} = 4 - 1 = 3

Point B(-3, 0)

x_{B} = 1\\ x_{B'} = 1 \\\\\frac{y_{B} +y_{B'} }{2} =2\\y_{B'} = 4 - y_{B} = 4 - 0 = 4

Point C(3, 4)

x_{C} = 1\\ x_{C'} = 1 \\\\\frac{y_{C} +y_{C'} }{2} =2\\y_{C'} = 4 - y_{C} = 4 - 4 = 0

Point D(-2, -1)

x_{D} = 1\\ x_{D'} = 1 \\\\\frac{y_{D} +y_{D'} }{2} =2\\y_{D'} = 4 - y_{D} = 4 - (-1) = 4 + 1 = 5

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Answer:

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Step-by-step explanation:

u may type it out in the comments...

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Let the number be x

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