# Thread: Clarifying on how to make a Sign Chart for Concavity + Max/Min

1. ## Clarifying on how to make a Sign Chart for Concavity + Max/Min

Hi,
I was just making sure if I was getting the hang of making a sign chart.

So it's basically critical numbers of the derivative, then you would take this derivative and put it into an x and f(x) sign chart.
After that you would plug in points between critical numbers to find where it ends up positive, and where it ends up negative, thus you'll be able to find the maximum if it goes from negative to positive (With the critical number in between, meaning that is the maximum point), and when it's positive to negative, then it would be a minimum.

For the Concavity,
It would just be Points of Inflections instead of critical numbers (which is similar to critical numbers to the first derivative, since it's the zeroes, but this time it's the zeroes for the second derivative)

So that would mean it's the same thing, use the points of inflections and see whether it concaves up and concaves down?
But just making sure, it's the same way right? From Positive to Negative is Concave Up, and Negative to Positive is Concave down?
Oh wait... isn't it the max and min as well, so I'm not sure whether it means concave up and concave down.

And Increasing would be both positive (as in positive critical number positive), and decreasing would be both negative.

So did I get the sign chart down right?
Just making sure if I got the hang of it. Thanks!

2. ## Re: Clarifying on how to make a Sign Chart for Concavity + Max/Min

A function is concave up if and only if its second derivative is positive. It is concave down if and only if its second derivative is negative. Points of inflection are points where the second derivative changes sign which means that "second derivative equals zero" is a necessary condition but not a sufficient condition. I don't know what sign chart you are talking about. You ask "So did I get the sign chart down right?" but show no sign chart.

3. ## Re: Clarifying on how to make a Sign Chart for Concavity + Max/Min

Suppose $f''$ is continuous on an open interval that contains $x = c$

1. If $f'(c) =0$ and $f''(c) < 0$ then $f$ has a local maximum at $x=c$

2. If $f'(c) =0$ and $f''(c) > 0$ then $f$ has a local minimum at $x = c$

3. If $f'(c) =0 \text{ and } f''(c) = 0$ and then the test fails. The function $f$ may have a
local maximum, a local minimum, or neither.

Here's an example. Hope it will help. Let's suppose there's a function:

$f(t) = 2t^3 - 14t^2 + 22t - 5, t \geq 0$

$\frac{d(f(t))}{dt} = 6t^2 -28t + 22$

\begin{align*}6t^2 -28t + 22 =& 0\\ t =& 1 \text{ or } t = \frac{11}{3}\end{align*}

Take any value between upper and lower limit of an interval for the following table.

 img.top {vertical-align:15%;} $\text{Interval }$ img.top {vertical-align:15%;} $0 < t < 1$ img.top {vertical-align:15%;} $1 < t < \frac{11}{3}$ img.top {vertical-align:15%;} $\frac{11}{3} < t$ img.top {vertical-align:15%;} $\text{Sign of } \frac{d(f(t))}{dt}$ img.top {vertical-align:15%;} $+$ img.top {vertical-align:15%;} $-$ img.top {vertical-align:15%;} $+$ img.top {vertical-align:15%;} $\text{Graph of f(t)}$ img.top {vertical-align:15%;} $\text{increasing }$ img.top {vertical-align:15%;} $\text{decreasing }$ img.top {vertical-align:15%;} $\text{increasing}$

$\frac{d(f(t))}{dt^2} = 12t -28$

Inflection point is at:

\begin{align*}12t -28 =& 0\\ t =& \frac{7}{3}\end{align*}
Again take any value between interval for the following table.

 Interval img.top {vertical-align:15%;} $0 < t < \frac{7}{3}$ img.top {vertical-align:15%;} $\frac{7}{3} < t$ img.top {vertical-align:15%;} $\text{Sign of } \frac{d(f(t))}{dt^2}}$ img.top {vertical-align:15%;} $-$ img.top {vertical-align:15%;} $+$ img.top {vertical-align:15%;} $\text{Graph of f(t)}$ img.top {vertical-align:15%;} $\text{concave down }$ img.top {vertical-align:15%;} $\text{concave up}$

So local maximum is at $t = 1$ because $\frac{d(f(1))}{dt} = 0$ and $\frac{d(f(1))}{dt^2} < 0$

and local minimum is at $t = \frac{11}{3}$ because $\frac{d(f(\frac{11}{3}))}{dt} = 0$ and $\frac{d(f(\frac{11}{3}))}{dt^2} > 0$

You can check it for yourself from the graph given below:

This example is taken from the book Thomas' Calculus Page-269-270