Sketching a of a graph with Calulus properties

[lamaclass]

I wanted to know if someone could check my graph and see if I'm on the right track with getting the correct graph. It has to have the following properties:

1. lim f(x) = oo; lim f(x) = -00
x --> 2[sup:3iu9oibc]+[/sup:3iu9oibc] x --> 2[sup:3iu9oibc]-[/sup:3iu9oibc]

2. lim f(x) = 0
x --> -00

3. f(5)=1; f(-2)=2; f(0)=0

4. f'(-2)=0; f'(5)=0

5. f'(x)>0 if x<-2 or x>5

6. f'(x)<0 if -2<x<2 or 2<x<5

7. f''(x)>0 if x <-3 or x>2

8. f''(x) <0 if -3<x<2

My graph:

 

[soroban]

Hello, lamaclass!

Sorry, you seemed to have missed most of the clues . . .

Sketch a graph with the following properties:

\(\displaystyle 1.\;\;\lim_{x\to2+}f(x) \:=\:\infty \qquad \lim_{x\to2^-}f(x) \:=\:-\infty\)

\(\displaystyle 2.\;\;\lim_{x\to-\infty}f(x) \:=\: 0\)

\(\displaystyle 3.\;\; f(5)\:=\:1,\;\; f(\text{-}2)\:=\:2,\;\; f(0)\:=\:0\)

\(\displaystyle 4.\;\; f'(\text{-}2)\:=\:0,\;\; f'(5)\:=\:0\)

\(\displaystyle 5.\;\;f'(x)\:>\:0\quad\text{if }((x<\text{-}2) \cup (x>5)\)

\(\displaystyle 6.\;\; f'(x)\:<\:0\quad\text{if }(\text{-}2<x<2) \cup (2<x<5)\)

\(\displaystyle 7.\;\; f''(x)\:>\:0\quad\text{if }(x <\text{-}3) \cup (x>2)\)

\(\displaystyle 8.\;\; f''(x)\:<\:0\quad\text{if }\text{-}3\,<\,x\,<\,2\)

Let's examine the properties one at a time . . .

\(\displaystyle 1.\;\;\lim_{x\to2+}f(x) \:=\:\infty \qquad \lim_{x\to2^-}f(x) \:=\:-\infty\)
There is a vertical asymptote at \(\displaystyle x = 2.\)
On the right, it "goes up". . On the left, it "goes down". .\(\displaystyle \searrow|\nwarrow\)

\(\displaystyle 2.\;\;\lim_{x\to-\infty}f(x) \:=\: 0\)
As we go to the far left, the graph approaches the x-axis.
(To the left, the x-axis is a horizontal asymptote.)

\(\displaystyle 3.\;\; f(5)\:=\:1,\;\; f(\text{-}2)\:=\:2,\;\; f(0)\:=\:0\)
We have three points: .\(\displaystyle (-2,2),\;(0,0),\;(5,1)\)

\(\displaystyle 4.\;\; f'(\text{-}2)\:=\:0,\;\; f'(5)\:=\:0\)
Horizontal tangents at \(\displaystyle x = \text{-}2\) and \(\displaystyle x = 5\)
\(\displaystyle (\text{-}2,2)\) and \(\displaystyle (5,1)\) are critical points (max or min).

\(\displaystyle 5.\;\;f'(x)\:>\:0\quad\text{if }(x<\text{-}2) \cup (x>5)\)
For \(\displaystyle x < \text{-}2\) and \(\displaystyle x > 5\), the function is increasing \(\displaystyle (\nearrow)\).

\(\displaystyle 6.\;\; f'(x)\:<\:0\quad\text{if }(\text{-}2<x<2) \cup (2<x<5)\)
For \(\displaystyle \text{-}2<x<2\) and \(\displaystyle 2<x<5\), the function is decreasing \(\displaystyle (\searrow)\)

\(\displaystyle 7.\;\; f''(x)\:>\:0\quad\text{if }(x <\text{-}3) \cup (x>2)\)
For \(\displaystyle x < \text{-}3\) and \(\displaystyle x > 2\), the function is concave up: \(\displaystyle \cup\)

\(\displaystyle 8.\;\; f''(x)\:<\:0\quad\text{if }\text{-}3\,<\,x\,<\,2\)
For \(\displaystyle \text{-}3<x<2\), the function is concave down: \(\displaystyle \cap\)

Note that the concavity changes at \(\displaystyle x = \text{-}3\)
There is an inflection point at \(\displaystyle x = \text{-}3.\)


The graph looks somethihng like this . . .

                         |    :*
                         |    :                 *
          (-2,2)-        |    : *              *
             o           |    :  *           *
         *        *      |    :    *      *
    *                 *  |    :       o(5,1)
 -------------------------o----:-------------------- 
                         | *  :2
                         |  * :
                         |    :

 

[lamaclass]

Thanks Soroban! I see it now! And it makes there'd be two graphs for it, makes much more sense!