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

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Answer 1 · 2016-12-18T02:13:02

See explanation and graph.

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]}

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)f(x)=x-x^(2/3)(5/2-x)?