Question

How do you sketch the graph by determining all relative max and min, inflection points, finding intervals of increasing, decreasing and any asymptotes given `f(x)=x-x^(2/3)(5/2-x)`?

Answer 1

See explanation and graph.

Explanation:

Rearranging,

`y=f(x)=(x^(5/3)+x-2.5x^(2/3))`

The graph passes through the origin (0, 0)

x-intercept ( y = 0 ): 1.4, nearly.

As `x to +-oo, y to +-oo`.

So, there are no asymptotes.

`y'=5/3x^(2/3)+1-(5/3)/x^(1/3)`

At x =0, y' has infinite discontinuity. It changes from `oo` to`-oo`.

and becomes 0 near x = 0.5, for sign change, from `-` to +

For `x < 0. y uarr, x in (0, 0.5), y darr, and x > 0,5` ( nearly), `y uarr`,

again .

There is a turning point near x = 0.5, for

the local minimum `= y(0.5)= -0.76`, nearly.,

,Relative maximum y = 0, at the cusp (0, 0).

The origin is a cusp and the tangent does not cross the curve,

there is no point of inflexion

graph{x^(2/3)(-2.5+x^(1/3)+x) [-2.5, 2.5, -1.25, 1.25]}