Can you tell me what the domain and range are for this function? 4x / x^2+1?

Question

Please explain in detail I want to understand. thanks

Answer 1

This function is defined for every real x. Note that x^2 + 1 >= 1 > 0 for every real x, so that it's denominator never vanishes. So, it's defined for every real x.

It's continuous, because it's given by the ratio of 2 continuous functions , that is, the functions x -> x, the identity function, and the function x -> x^2 + 1, which never vanishes on R. In addition, it's an odd function, that is f(-x) = - f(x)

We see that f(x) = 0 and that lim x-> oo f(x) = lim x -> oo 4x / (x^2+1) = 0 (the denominator goes to infinity faster than the numerator). Now let's find the extermum points of f. So, lets's compute it's derivative f' (f is differentiable because it's give by the ratio of the 2 differentiable functions given above). According to the rule for diferentiating a the qupciente of 2 differentiable functions, it folllows that

f'(x) = 4 [ (x^2 + 1) - x(2x)]/(x^2 + 1)^2 = 4(1 - x^2)/(x^2 + 1)^2. Therefore, f' vanishes at x' = 1 and at x" = -1. We see that, on the positive real axis, f'(x) > 0 for x < 1and f'(x) < 0 for x >1, which implies f has a maximum at x' = 1. In addition, this implies f is increasing in [0 ,1) and decreasing in (1, oo), so that it has a global maximum on the positive real axis at x' = 1. The maximum is therefore f(1) = 4/2 = 2.

Since f is odd, a similar reasoning shows it has a global minimum on the negative real axis at x" = -1, for which f(-1) = -2. We conclude x" = -1 is a global minimum on the entire real axis. Since f is continuous, it follows it takes on every value on [-2, 2] which is it's range.

Answer 2

The domain is certainly all real numbers because the function is defined for all real x. However, the range is not all real numbers. When the denominator is a polynomial of higher power than the numerator there will always be limiting values on the function. You need to differentiate and find the x values for maximum and minimum and then put these values into the function to find maximum y and minimum y. The range is between these points including them. You should be able to see that as x gets very large positive or negative the fraction will get very small and tend toward zero. If you work through it you will find that the range is between two small integers.

Answer 3

A domain is the set of values that can be placed into the equation that will make the results of the equation real numbers and thus valid. In this example, the domain of the equation is all real numbers, basically from negative infinity to infinity. When we look at the numerator, for any value you plug into x, you will have a valid result (positive, negative, or zero). The bigger constraint of domain is in the denominator of this equation, where the result there must not equal 0, or else it isn't real. Now, the only way for the denominator to equal 0 is for x^2 to be equal to -1. Since we cannot square any real number positive or negative to equal -1, we can say that the domain for the entire function is from negative infinity to infinity, since any value is valid.

The range of a function is basically the values (results) of the function after plugging in values for x. We look at the domain limits as well as possible areas of discontinuity and we can look at what happens to the result if we were to plug these in. If we plug in infinity, the denominator has a much higher order (x^2 > x), and it will grow much faster in size than the numerator. As a result the values obtained will eventuall equal zero. The same goes for negative infinity. The most important part of this function is therefore around the area where x is equal to 0. If we were to graph this, we would see that the function is at its highest when x is equal to 1 and lowest when it is equal to -1. The results of this function when we plug in those values gives us 2 and -2, thus giving us our range.

Answer 4

The domain is real numbers and the range is also real numbers. But in concerned in other functions because some functions are discountinous, meaning the value of denominator is zero(which is undefine). U should not set the value of the domain to become the the range is undefine. To determine what is that value, you should equate the root is equal to zero. If the result is real no. therefore the function is discontinous function. (That only occur in fractional equation).

Answer 5

the domain and range are real numbers

Answer 6

be conscious that f(x)=sqrt(4-x^2)/sqrt(a million-x^2) is defined for all x such that x^2<=4 and x^23 so (4-x^2)/(a million-x^2) >4 and its sq. root is >2 for that reason the style is contained in (2,infinity) to coach it is precisely (2,infinity), shall we desire to coach that for any y>2, we are able to locate an x such that sqrt(4-x^2)/sqrt(a million-x^2)=y it is finished by fixing for x sq. the two aspects (4-x^2)/(a million-x^2)=y^2 i.e., a million+3/(a million-x^2)=y^2 3/(a million-x^2)=(y^2-a million) (a million-x^2)=3/(y^2-a million) x^2=a million-3/(y^2-a million)=(y^2-4)/(y^2-a million) so we are able to take x=sqrt(y^2-4)/sqrt(y^2-a million) once you think approximately that y^2-4>0 and y^2-4=x^2, i.e., x>=0 and x<=4 So section is [0,4] style is [0,infinity] on the grounds which you will possibly be able to desire to freshen up sqrt(4x-x^2)=y for any y>=0 {i bypass to bypass away it so which you will freshen up]

Answer 7

domain and range all real x

Answer 8

I really do not know