Local maximum/minimum of a cubic function

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March 2nd 2010, 11:54 AM
Jools

Local maximum/minimum of a cubic function

I have a question that gives me the derivative graph (but not the function itself) of a cubic Function. The g '(x) is a parabola and is attached below:

Given the graph of g '(x), sketch g "(x) and a possible graph of g(x). Identify the following:

a) The interval where g(x) is increasing and decreasing

b) The local maximum and minimum points of g(x)

c) The intervals where g(x) is concave up and concave down

I have figured out the intervals where g(x) is increasing and decreasing as:

g '(x) > 0 when x < -5 and x > 1, therefore, g(x) is increasing

g '(x) < 0 when -5 < x < 1, therefore, g(x) is decreasing.

I have also attached what I have graphed as g(x) and g "(x).

Could anybody help me with finding the local maximum and minimum of g(x). I'm kind of stuck...

http://i299.photobucket.com/albums/m...976/maxmin.jpg

http://i299.photobucket.com/albums/m...2010_00000.jpg
March 2nd 2010, 01:48 PM
Soroban

Hello, Jools!

Quote:

I have a question that gives me the derivative graph of a cubic function.
The graph of $g'(x)$ is a parabola and is attached below.

Given the graph of $g '(x)$ , sketch $g''(x)$ and a possible graph of $g(x).$

Identify the following:

a) The intervals where $g(x)$ is increasing and decreasing

b) The local maximum and minimum points of $g(x)$

c) The intervals where $g(x)$ is concave up and concave down

I have figured out the intervals where g(x) is increasing and decreasing as:

. . $\begin{array}{cccc}g'(x) > 0 &\text{when }x < -5\,\text{ or }\,x > 1 & g(x)\text{ is increasing} \\<br /> <br /> g '(x) < 0 & \text{when }-5 < x < 1 & g(x)\text{ is decreasing} \end{array}$ . Right!

I have also attached what I have graphed as $g(x)$ and $g''(x)$ . Correct!

With the given points of the parabola: . $(-5,0),(0,-5), (1,0)$
. . we can determine its equation: . $g'(x) \:=\:x^2+4x-5$

Then: . $g(x)\;=\;\int(x^2+4x-5)\,dx \;=\;\tfrac{1}{3}x^3 + 2x^2 - 5x + C$

The extreme values occur where $g'(x) = 0 \quad\Rightarrow\quad x \:=\:-5,\:1$

The second derivative is: . $g''(x) \:=\:2x + 4$

. . $\begin{array}{cccccccc}g''(\text{-}5) &=& 2(\text{-}5)+4 &=& -6 & \text{concave down: }\cap & \text{maximum} \\ g''(1) &=& 2(1)+4 &=& +6 & \text{concave up: } \cup & \text{minimum}\end{array}$

Therefore: . $\begin{Bmatrix}\text{maximum at:}& \left(-5,\:C + \frac{100}{3}\right) \\ \\[-3mm] \text{minimum at:} & \left(1,\:C-\frac{8}{3}\right) \end{Bmatrix}$

Inflection points: . $g''(x) \:=\:0 \quad\Rightarrow\quad 2x + 4 \:=\:0 \quad\Rightarrow\quad x \:=\:-2$

Therefore: . $\begin{Bmatrix}g''(x) > 0 & \text{when }x > -2 & \text{concave up} \\ \\[-3mm] g''(x) < 0 & \text{when }x < -2 & \text{concave down} \end{Bmatrix}$
March 2nd 2010, 02:26 PM
Jools

Quote:

Originally Posted by Soroban

Hello, Jools!

With the given points of the parabola: . $(-5,0),(0,-5), (1,0)$
. . we can determine its equation: . $g'(x) \:=\:x^2+4x-5$

Then: . $g(x)\;=\;\int(x^2+4x-5)\,dx \;=\;\tfrac{1}{3}x^3 + 2x^2 - 5x + C$

The extreme values occur where $g'(x) = 0 \quad\Rightarrow\quad x \:=\:-5,\:1$

The second derivative is: . $g''(x) \:=\:2x + 4$

. . $\begin{array}{cccccccc}g''(\text{-}5) &=& 2(\text{-}5)+4 &=& -6 & \text{concave down: }\cap & \text{maximum} \\ g''(1) &=& 2(1)+4 &=& +6 & \text{concave up: } \cup & \text{minimum}\end{array}$

Therefore: . $\begin{Bmatrix}\text{maximum at:}& \left(-5,\:C + \frac{100}{3}\right) \\ \\[-3mm] \text{minimum at:} & \left(1,\:C-\frac{8}{3}\right) \end{Bmatrix}$

Inflection points: . $g''(x) \:=\:0 \quad\Rightarrow\quad 2x + 4 \:=\:0 \quad\Rightarrow\quad x \:=\:-2$

Therefore: . $\begin{Bmatrix}g''(x) > 0 & \text{when }x > -2 & \text{concave up} \\ \\[-3mm] g''(x) < 0 & \text{when }x < -2 & \text{concave down} \end{Bmatrix}$

Thanks for the response! Is there a way to determine the max/mins without working out the actual equation? I'm pretty sure that's what the question in my book is asking me to do.... By looking at the graph wouldn't the max be (-5,0)? and the min be (1,??)? Thanks!