exponential graph

Question

Is there an easier way of getting my values for f(x)=x^2/3 

I've done it like this: divided 2/3 and raised whatever x value i took to the result of my division. 

Example 8^2/3 = 3.999 and rounded it to 4. That was my primitive way of doing it, im sure there is more mathematical approach. 

 

Answer 1

Hi, Take Arturo's advice! 

 

more generally, when you have fractional exponents, in most cases you will make your life more difficult if you change them into decimals. This is especially true when the number has to be approximated by rounding or truncating, as with 2/3.

 

You're better off considering the exponent as an exponent and a root separately... the denominator is the root, and the numerator is the exponent.

 

So for graphing, build a table using input values of x that are perfect cubes, like the 8 you already picked. Then you take the cube root first, and square the result.

 

So... 3 cubed is 27. Consider x=27... f(x) = (27^1/3)^2  = 3^2 = 9

 

2 cubed is 8. f(8) = (8^1/3)^2 = 4

 

f(1) is pretty easy... it's 1

 

try using 4 cubed... that's 64, f(64) = 16.

 

Now use the negative values -1, -8, -27, -64 and see what you get for those.

 

Sometimes the easiest way to do a problem is to figure out which input values are the easiest to figure out!

 

--Louise K

Answer 2

This can written as

 

f(x) = ³√(x²)

 

 

The domain for a cube-root function is all real numbers because you can take the cube-root of negative and positive numbers.  In interval notation, (-∞ , ∞).

 

Next, we can see how the f(x) act as x reaches infinity and negative infinity.  This will give you an idea of the shape of the graph as well as any limits of f(x) values.

 

If we evaluate f(x) values in the interval (0, ∞),

 

f(1) = ³√1 =

f(10) = ³√(100) = 4.64

f(100) = ³√(10000) =  21.54

f(10000) = ³√(100000000) = 464.16

 

 

As you can see, the f(x) value increase infinitely.

 

If we evaluated f(x) at x=-1 , x=-10 , x=-100, and x=-10000, the f(x) values reflect upon themselves with respect to the positive x values we just evaluated f(x) for.

 

Because you taking the square of some number, the result is always positive.  Taking the cube-root of a positive number is always positive.  So your range is [0, ∞).

 

 

This method is more efficient, rather than evaluating f(x) at every single number.

 

Answer 3

Rick, remember that if we have a power to a power, we multiply the exponents, like (x²)³  = x⁶.

 

Remember also that the square root of x is also written as x^1/2.

 

So, x^2/3 = (x^1/3)² or (x²)^1/3

 

8^2/3 = (8^1/3)² = 2² = 4 or (8²)^1/3 = 64^1/3 = 4.

 

Hope this helps.