# Graph sketching

## Homework Statement

I know for graphs of function f(x)=x^n where n is an odd power, even power or square root have their own pattern but how about

f(x)=x^(3/2)

or f(x)=x^(1/2)

is that considered odd or even ?

## The Attempt at a Solution

Related Precalculus Mathematics Homework Help News on Phys.org
Gib Z
Homework Helper
No, 3/2 and 1/2 are neither even or odd. There isn't exactly a simple rule for f(x) = x^(3/2), but it is a famous curve called http://mathworld.wolfram.com/SemicubicalParabola.html" [Broken]

You can think of f(x) = x^(1/2) as a parabola tipped on its side, and then the bottom half is chopped off so that it's a function.

Last edited by a moderator:
No, 3/2 and 1/2 are neither even or odd. There isn't exactly a simple rule for f(x) = x^(3/2), but it is a famous curve called http://mathworld.wolfram.com/SemicubicalParabola.html" [Broken]

You can think of f(x) = x^(1/2) as a parabola tipped on its side, and then the bottom half is chopped off so that it's a function.
thanks, i have tried graphing several such graphs with a program and notice something.

Any function f(x)=x^n , where n is 1/2, 1/3 (the denominator can be any real and the numerator is 1), the graph will look like a one-sided parabola opening to the right

And if n=3/2, 5/2, 7/3 (any rational numbers aside from case 1)

the graph will look like a semicubical parabola.

Last edited by a moderator:
eumyang
Homework Helper
thanks, i have tried graphing several such graphs with a program and notice something.

Any function f(x)=x^n , where n is 1/2, 1/3 (the denominator can be any real and the numerator is 1), the graph will look like a one-sided parabola opening to the right
Err, not quite. The graph of $$f(x) = x^{1/3}$$ does NOT look like a one-sided parabola opening to the right.

The function $$f(x) = x^{2}$$ is a parabola, but it is not one-to-one. If we restrict the domain of f(x) to [0, ∞), then f(x) would be one-to-one and the inverse would be $$f^{-1}(x) = x^{1/2} = \sqrt{x}$$. So the graph of $$f^{-1}(x)$$ would be a half-of-a-parabola laying on its side.

Now, the function $$g(x) = x^{3}$$, your basic cubic, IS one-to-one, so we don't need to restrict the domain. It's inverse would be $$g^{-1}(x) = x^{1/3} = \sqrt[3]{x}$$, and its graph would look like the COMPLETE graph of $$g(x) = x^{3}$$, but rotated to the side and flipped, for a lack of a better desciption.

69

Err, not quite. The graph of $$f(x) = x^{1/3}$$ does NOT look like a one-sided parabola opening to the right.

The function $$f(x) = x^{2}$$ is a parabola, but it is not one-to-one. If we restrict the domain of f(x) to [0, ∞), then f(x) would be one-to-one and the inverse would be $$f^{-1}(x) = x^{1/2} = \sqrt{x}$$. So the graph of $$f^{-1}(x)$$ would be a half-of-a-parabola laying on its side.

Now, the function $$g(x) = x^{3}$$, your basic cubic, IS one-to-one, so we don't need to restrict the domain. It's inverse would be $$g^{-1}(x) = x^{1/3} = \sqrt[3]{x}$$, and its graph would look like the COMPLETE graph of $$g(x) = x^{3}$$, but rotated to the side and flipped, for a lack of a better desciption.

69
yeah that's only when the denominator is odd. Thanks.