Curve Sketching and Asymptotes Question

ardentmed

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For \(\displaystyle f(x)\, =\, 3x^{\frac{2}{3}}\, -\, x,\) give the intervals of increase and decrease, identify any local extrema, give the intervals where concave up and concave down, and the coordinates of any points of inflection. Identify any asymptotes and sketch the curve.
Here is the result I obtained:

So there is no horizontal asymptote nor is there a vertical asymptote since no x value will result in an undefined y value.

Moreover, the domain is (-infinity, 0)u(0,infinity) due to the cusp at x=0. Moreover, f'(x) and f''(x) are both yield DNE at x=0.

There should be a local maximum at f(8) = 4 due to the increasing slope to the left and decreasing slope to the right.

There is no symmetry since f(x) =/ f(-x)

Thus, the aforementioned information should result in this graph:

snapshot (2).jpg

Am I on the right track?

Thanks in advance.
 
For \(\displaystyle f(x)\, =\, 3x^{\frac{2}{3}}\, -\, x,\) give the intervals of increase and decrease

What do you get, for these intervals?


identify any local extrema

There should be a local maximum at f(8) = 4 due to the increasing slope to the left and decreasing slope to the right.

That's correct. The point (8,4) is a local maximum.


give the intervals where concave up and concave down, and the coordinates of any points of inflection.

What is your answer, for these?


Identify any asymptotes and sketch the curve.

So there is no horizontal asymptote nor is there a vertical asymptote

View attachment 4305

You're correct -- there are no asymptotes.

Your blurry graph-image looks like it's flipped horizontally.

(The local maximum belongs in Quadrant I.)



the domain is
(-infinity, 0)u(0,infinity) due to the cusp at x=0

That's the domain of the function's derivatives.

You already wrote that there's no value of x causing f(x) to be undefined, yes? This is true, and it means that the domain of function f is all Real numbers.

I cannot determine from your image whether you plotted an open circle at (0,0). If you did, fill it in. The point (0,0) needs to be included, on your graph.


There is no symmetry since f(x) =/ f(-x)

That's the test for Even symmetry only.

There is a different test for Odd symmetry. Yet, you're correct. Function f has no symmetry.

Cheers :)
 
Last edited:
How do you do the odd symmetry test?

As for concavity, f is concave down from (-infinity, infinity)

f is decreasing from (-infinity,0)u(8,infinity) and increasing frolm (0,8)

No infection points exist.

Thanks again. Also, yes, my webcam "mirrored" the image by accident.

Cheers.
 
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