Question

How do you find the points where the graph of the function `f(x) = (x^3 + 2) / (x^(1/3))` has horizontal tangents and what is the equation?

Answer 1

Horizontal tangent at `x=1/4^(1/3)`, equation of horizontal tangent is `y=9/4^(8/9)`

Explanation:

First, simplify `f(x)` by splitting up the fraction.

`f(x)=x^3/x^(1/3)+2/x^(1/3)`

`f(x)=x^(8/3)+2x^(-1/3)`

In order to find the horizontal tangents of a function, we must find the times when the derivative of the function equals `0`.

To differentiate `f(x)`, use the power rule.

`f'(x)=8/3x^(5/3)-2/3x^(-4/3)`

It will be easier to find when this equals `0` if we put it into fractional form. First, simplify by putting the `3` in the denominator of a combined fraction.

`f'(x)=(8x^(5/3)-2x^(-4/3))/3`

Now, multiply the fraction by `x^(4/3)/x^(4/3)`.

`f'(x)=(8x^3-2)/(3x^(4/3))`

We can now set this equal to `0` to find the original function's horizontal tangents.

`(8x^3-2)/(3x^(4/3))=0`

`8x^3-2=0`

`x^3=1/4`

`x=4^(-1/3)`

However, the line of the horizontal tangent will be in the form `y=ul(" ")`. Thus, we must find the function value at `x=4^(-1/3)`.

`f(4^(-1/3))=((4^(-1/3))^3+2)/((4^(-1/3))^(1/3))=(1/4+2)(4^(1/9))=9/4(4^(1/9))=9/(4^(8/9))`

The horizontal tangent is the line `y=9/4^(8/9)`.

Graphed are the function and its tangent line:

graph{((x^3 + 2) / (x^(1/3))-y)(y-9/4^(8/9)-0.0001x)=0 [-19.65, 20.9, -8.3, 12]}