What are the domain and range of the real-value function f(x) =-2+√x+7

Mathematics

2 answers:

prohojiy [21]3 years ago

8 0

The given function -2+√x+7 is a square root function.

The expression inside the square root of a unction cannot be negative. Because if it will be negative then the function will be imaginary. So, expression inside the square root must be positive or zero. Hence, to find the domain of a square root function set the expression inside the square root as more than or equal to zero.

So, x+7≥0

x+7 -7 ≥0-7 Subtract 7 from each sides to isolate x.

x≥ -7

So, the domain of the given function is x≥ 7.

Notice that if we will plug in any value which is more than or equal to 7then we will get y >=-2 only. It must be -2+.

So, range of this function is y>=-2.

stiks02 [169]3 years ago

4 0

Because the function f(x) =-2+√x+7 has √x, x should be more or equal 0.

Domain is x≥0, or [0,+∞).

The smallest value of x=0, so the smallest value of f(x) is f(0) =-2+√0+7=5,

so range is f(x)≥5, or [5, +∞).

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Let n be a positive integer and define [n] to be the set of the first n positive integers. That is, [n] = {1, 2, 3, . . . , n}.

yaroslaw [1]

Answer: There are $2^{n-1}$ ways of doing this

Hi!

To solve this problem we can think in term of binary numbers. Let's start with an example:

n=5, A = {1, 2 ,3}, B = {4,5}

We can think of A as 11100, number 1 meaning "this element is in A" and number 0 meaning "this element is not in A"

And we can think of B as 00011.

Thinking like this, the empty set is 00000, and [n] =11111 (this is the case A=empty set, B=[n])

This representation is a 5 digit binary number. There are $2^5$ of these numbers. Each one of this is a possible selection of A and B. But there are repetitions: 11100 is the same selection as 00011. So we have to divide by two. The total number of ways of selecting A and B is the $2^{5-1} = 2^4$ .