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November 14th, 2015, 05:25 AM   #1
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Critical numbers, concavity and inflection points

I really need help with this problem. I've been trying for hours, and I don't really understand this topic. If you could show me how to do this, and explain, then I would greatly appreciate it! Thank you!

f(x) = (1-x)(x+3)^2

A) What interval is it increasing and decreasing on?
B) Find the x-coordinates of all local maxima of f, compute their average.
C) Find the x-coordinates of all local minima of f, compute their average.
D) Use interval notation to indicate where f(x) is concave up and concave down.
E) Find all inflection points of f, compute their average.

Last edited by skipjack; November 14th, 2015 at 07:12 AM.
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November 14th, 2015, 05:31 AM   #2
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Where in the world did you get this problem? It is a fairly standard Calculus problem but you seem to be saying that you have not taken Calculus. Is that correct? Do you know what the derivative of a function is? Do you know what it means when the derivative is 0? When it does or does not change sign?
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November 14th, 2015, 05:36 AM   #3
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I'm in the process of taking the class now. This is the current topic that we are on, but my professor doesn't really speak English, so basically it's a "teach it to yourself class." I tried going to office hours with this, and he was no help then either. I can calculate the derivative of the problem, and I calculated out the critical points, but then I got stuck.
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November 14th, 2015, 08:07 AM   #4
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As f$\,'\!$(x) is negative for x < -3, positive for -3 < x < -1/3, and negative for x > -1/3,
f(x) has a local minimum at (-3, 0) and a local maximum at (-1/3, 256/27).

As f$\,''\!$(x) = -6x - 10, the point (-5/3, 128/27) is the only point of inflection of f(x).
Its coordinates may be found by averaging the coordinates of the local minimum and the local maximum.

The function f(x) is concave up on (-$\infty$, 128/27] and concave down on [128/27, $\infty$).
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November 14th, 2015, 08:20 PM   #5
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Quote:
Originally Posted by skipjack View Post
As f$\,''\!$(x) = -6x - 10, the point (-5/3, 128/27) is the only point of inflection of f(x).

The function f(x) is concave up on (-$\infty$, 128/27] and concave down on [128/27, $\infty$).

No, the function f(x) is concave up on (-$ \infty$, -5/3) and concave down on (-5/3, $\infty$).
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November 14th, 2015, 11:32 PM   #6
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Oops, thanks for the correction - I copied the function's value instead of the value of x.
However, that value should be included by using square brackets: concave up on (-$\infty$, -5/3] and concave down on [-5/3, $\infty$).
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November 15th, 2015, 10:25 AM   #7
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Quote:
Originally Posted by skipjack View Post
Oops, thanks for the correction - I copied the function's value instead of the value of x.
However, that value should be included by using square brackets: concave up on (-$\infty$, -5/3] a
nd concave down on [-5/3, $\infty$).

No, that value should not be included.

"A function $y = f(x)$ is concave up on an open interval if $y''$ is positive on the interval.
And a function $y = f(x)$ is concave down on an open interval if $y''$ is negative on the interval."

Source:

http://www.millersville.edu/~bikenag...conc/conc.html



They have to be open intervals. At x = -5/3, the value of the second derivative is zero.


.
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Last edited by Math Message Board tutor; November 15th, 2015 at 10:32 AM.
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November 15th, 2015, 11:24 AM   #8
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Nice try, but here's another source that clearly supports my views.

Math Tutor - Derivatives - Theory - Graphing

My view is also supported on wikipedia, which gives a definition that doesn't require that the function is twice differentiable. I prefer this approach. Consider the function $y = |x|^{4/3}$. It certainly "looks" concave up everywhere, but is not twice differentiable at $x$ = 0.

At least one other source, though, states "there seems to be no general agreement on a definition".
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