What is the range of the function y=^3 sqrt x+8

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In this problem, it wants us to find the range of this function. Um so in a cube root function right, it kind of looks like this for the answers are going to go on forever to the left and right, since you can take the keyboard of any number and the same with up and down. So the range here is going to be all real numbers, from negative infinity to positive.

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To find the range of the function y =^3 Square root x+ 8, we need to find the function's domain and range. The domain of the function is all real numbers, while the range is -8, 8.

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Video Transcript

In this problem, it wants us to find the range of the function. Since you can take the keyboard of any number and the same with up and down, it looks like the answers will go on forever to the left and right in a cube root function. The range is going to be all real numbers.

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Algebra

Find the Domain and Range cube root of x-8

Step 1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

Set-Builder Notation:

Step 2

The range is the set of all valid values. Use the graph to find the range.

Interval Notation:

Set-Builder Notation:

Step 3

Determine the domain and range.

Domain:

Range:

Step 4

You can make a table of values for #f(x)# for various values of #x#
Remember that our functions is defined for #x>=8#

Hence its graph is

You can go to the link below where i have setup a desmos page to play with various values of f(x) which are then plotted so you get a feeling of the whole process.

desmos files for plotting f(x)

Want to join the conversation?

• If g(x)=x^2/x
  and
  x^2/x=x
  then can we say that:
  g(x)=x
  And if we can, how than for x=0 this expression is undefined?

  In Serbia, we learned that expressions needs to be simplified if it is possible. So x^2/x is same as x, and regular way to describe it is to write it like that, just x.

  I`m interested in this, because if this is the case with functions, any expression (X) could be writen like that - X^2/X which then means that no expression (X) is defined for X=0, regardless of what expression is.

  For example, F(x)=345x is defined for input value of 0, and has output of 0. But 345x=(345x)^2/345x, it is undefined for x=0.

  Im sorry if this is stupid question, and if i missed some basic rule with functions, but this is not clear to me. Can we transform expressions in functions like we do in other mathematical expressions or can we not?

  • When you start with a reciprocal function, you will have at least one vertical asymptote in which the function does not have a value. So by starting with g(x) = x^2/x, you have a vertical asymptote at x=0, so from the start of your problem, x cannot equal to zero. So when you reduce it to g(x)=x, you have to say that x=0 is an extraneous solution (see https://en.wikipedia.org/wiki/Extraneous_and_missing_solutions for definition of extraneous solution). Hope this answers your question.

• OK so I'm totally lost by this. Is this at all like the domain on the function? They seem totally different but also like the exact same thing.

  • 

    The domain of a function is the set of all acceptable input values (X-values).
    The range of a function is the set of all output values (Y-values).
    Hope this helps.

• At

  2:22

  Sal says that the definition is F(x) is going to be equal to x^2. Does that mean that if I have a function notation such as f(x)=x+4 and a given x is 2, do I have to square the 2?

  • 
    

    Yes - that is how it works, if you have f(x)=x² and are asked what is f(2), then you replace every instance of x in the function definition with 2 so given f(x) = x², that means f(2) = 2² = 4.
    Here is another example: If f(x) = x² + 5x then f(2) = 2² + (5)(2) = 4 + 10 = 14

• what is the difference between f(x) and y values?

  • 

    f(x) is Y.
    The equation f(x) = 2x - 5 is the same a y = 2x - 5 provided the equation with Y is a function. Not all equation are functions. So, you can't always swap out Y for f(x). However, you can always swap out f(x) for Y.

• Sorry if this was asked already but what is a non real number and what would an example be?

  • Non-real numbers are called imaginary numbers and are based on i=√-1. You cannot take the square root of negative numbers, so you move to imaginary number.

• What if the input is an imaginary number (i) ? how will the graph be?

• At

  3:43

  why cant y be negative? Is this true for all problems or only dis one

  • To answer you question of if it is always true, the answer is no. If you start from the quadratic parent function, y=x^2, then y cannot be negative.
    One way to include negatives is to reflect it across the x axis by adding a negative y = -x^2. With this y cannot be positive and the range is y≤0. The other way to include negatives is to shift the function down. So y = x^2 -2 shifts the whole function down 2 units, and y ≥ -2.

• i just wanna ask that we know that f(x)=y but is it true for variable other than x, is f(n)=y
  is this correct

  • We use x and y because they are graphed on a Cartesian Plane. f(n) is a little confusing because this is somewhat reserved for sequences where n is a set of whole numbers which most of the time start at 1, so that is not a good choice - saying f(n) = 3n + 1 would not be the same as y = 3n + 1. I suppose if you say that y = 3a + 5 (which could be done because a variable is just a variable), then saying f(a) = 3a + 5 would be equivalent, but is obscured by the inability to graph on a normal Cartesian Plane and possibly confusing to other mathematicians.

• Hi,

  I have a function f(x) = 1/(x-2) where x belongs to R. We know that

  Domain will be all numbers but 2 since the function will not be defined at 2.

  And the range is all real numbers except 0.

  The definition of range is the set of all possible values that the function will give when we give in the domain as input.
  My question is this:
  now if I were to use any value from the domain (say 50000 ) 1/x-2 will give me a value less than 1 i.e 1/49998. I am not able to find any value ,in the domain,which when substituted gives me a value greater than one. So shouldn't the range be (-1,1]?

  • The numerator is 1, so for the function value to be greater than 1, the denominator must be between 0 and 1.

    0 < 𝑥 − 2 < 1 ⇒ 2 < 𝑥 < 3

    Likewise,
    −1 < 𝑥 − 2 < 0 ⇒ 1 < 𝑥 < 2
    will yield a function value less than −1

• I don't understand the difference in the two equations represented other than one is f(x) and the other is g(x). How did you come up with two separate answers for the domain?

  • They are almost the same, but g(x) has to exclude 0 from the beginning because you cannot divide by 0 which is why Sal put a circle on the graph and f(x) includes 0. They are two different functions f(x) is quadratic and g(x) ends up almost linear. That is why their ranges are so different.

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