increasing/decreasing: f(x) = 3x^(2/3) - 2x

[jwpaine]

For -1 <= x <= 1, let f(x) = 3x^(2/3) - 2x

find the global extreme values, and the intervals on which f is increasing, decreasing, concave up, concave down. Does the graph of y = f(x) have an inflection?

\(\displaystyle \L f'(x) = \frac{2}{x^{1/3}} - 2\)

\(\displaystyle \L f'(x) = \frac{2}{x^{1/3}} - 2 = 0\), critical at: \(\displaystyle \L x = 1\)

end points:

\(\displaystyle \L f(-1) = 5\)

\(\displaystyle \L f(1) = 1\)


\(\displaystyle \L f''(x) = \frac{-2}{3x^{4/3}}\)

\(\displaystyle \L f''(x) = 0:\)undefined

Now how do I prove on what intervals the function is increasing/decreasing?

Graphically, I can see that x is decreasing on \(\displaystyle \L (-\infty\,\, ..\,\, 0)\), increasing on \(\displaystyle \L (0 \,\,..\,\, 1)\) and then decreasing on \(\displaystyle \L (1\,\,..\,\,+\infty)\)

John

 

[stapel]

Now how do I prove on what intervals the function is increasing/decreasing?

The number line is divided into three intervals by the one x-value at which the derivative is undefined and the other x-value where you have a critical point. Check the sign of the derivative on each of these intervals. (I'd check at x = -1 and at x = 1/8 = 1/2³.) If f' is positive, then f is increasing; if f' is negative, then f is decreasing.

Eliz.

 

[jwpaine]

Oh, ok, of course: finding the slope of the tangent at a point to see if it is increasing or decreasing.

f'(-1) = -4: decreasing --> -oo < x < 0
f'(1/8) = 2: increasing --> 0 < x < 1


Now finding the global max on this closed interval [-1 .. 1]: it would be my end point of x = -1, because f(-1) is the largest out of the three tested points (end points and critical), correct?

Now for concavity: I'll use the second derivative of f(x) to see how f'(x) is changing: (what is the point in this? This kind of goes hand-in-hand with the increasing/decreasing part I just did, doesn't it?)

So I assume I would test points of min/max on this closed interval:

f''(-1) > 0: convex
f''(1) < 0: concave

What about x = 0? x = 0 is undefined, so would this mean that it is not a point of inflection? Slope does change at x = 0... hmmmm

 

[Deleted member 4993]

Oh, ok, of course: finding the slope of the tangent at a point to see if it is increasing or decreasing.

f'(-1) = -4: decreasing --> -oo < x < 0
f'(1/8) = 2: increasing --> 0 < x < 1


Now finding the global max on this closed interval [-1 .. 1]: it would be my end point of x = -1, because f(-1) is the largest out of the three tested points (end points and critical), correct?

Now for concavity: I'll use the second derivative of f(x) to see how f'(x) is changing: (what is the point in this? This kind of goes hand-in-hand with the increasing/decreasing part I just did, doesn't it?)

So I assume I would test points of min/max on this closed interval:

f''(-1) > 0: convexf"(x) is always negative for the given problem...edited
f''(1) < 0: concave

You have a cusp at x = 0.


What about x = 0? x = 0 is undefined, so would this mean that it is not a point of inflection? Slope does change at x = 0... hmmmm

 

[jwpaine]

How so? The second derivative is: f''(x) = -2/(3x^(4/3))

f''(1) yields a negative, right?

 

[stapel]

f''(1) yields a negative, right?

That's what I'm getting, too. And the graph certainly looks as though f(x) is concave down at all times.

Eliz.

 

[Deleted member 4993]

Sorry - missed the leading negative sign.

However, the point is though that the sign of f"(x) does not change - it remains always negative. You have two "convex up" (or concave down - sad faces) curves joined at x=0.

 

[jwpaine]

Thank you very much.

I messed up on my arithmetic: f''(-1) and f''(1) are both less than zero.

John