# Thread: Calculus Graphing

1. ## Calculus Graphing

I am having trouble with this problem and asking if someone can show me a this step by step so i can do good on my test on Friday. THANKS

Sketch the graph of a continuous function f that satisfies all of the stated conditions.

f(0)=4; f(-3)=f(3)=0; f^l(-3)=0; f^l(0)is undefined;
f^l(x) >0 if -3 <x<0; f^l(x) <0 if x <-3 or x >0;
f^ll(x) > 0 if x <0 or <x <2; f^ll(x)<0 if x >2

2. Hello, uniquereason81!

Sketch the graph of a continuous function $\displaystyle f(x)$ such that:

. . $\displaystyle f(0)=4,\;\; f(\text{-}3)=f(3)=0\;\;{\color{blue}[1]}$

. . $\displaystyle f'(\text{-}3)=0,\;\;f'(0)\text{ is un{d}efined }\;{\color{blue}[2]}$

. . $\displaystyle f'(x) \:=\:\bigg\{\begin{array}{cc} > 0 & \text{if }-3 <x<0 \\ <0 & \text{if }x <-3\text{ or }x >0 \end{array}\;\;{\color{blue}[3]}$

. . $\displaystyle f''(x) \:=\:\bigg\{\begin{array}{cc}> 0 & \text{if }x <0\text{ or }0<x <2 \\ <0 & \text{if }x >2 \end{array}\;\;{\color{blue}[4]}$

From [1], we have three points: .$\displaystyle (\text{-}3,0),\0,4),\3,0)$

From [2], there is a horizontal tangent at (-3,0).
. . And the graph "goes vertical" at x = 0.
Code:
                |
o
|
|
|
--=o=--------+---------o--
-3         |         3
|

From [3], $\displaystyle f(x)$ is "uphill" for $\displaystyle -3 < x < 0$
. . and "downhill" everywhere else.
Code:
                |
*
\             / | \
\        /    |    \
\   /       |       \
------*---------+---------*--
-3         |         3 \
|             \

From [4], $\displaystyle f(x)$ is concave down for $\displaystyle x > 2$
. . and concave up everywhere else.
Code:
                |
*              *
*|*
*          *  |    *
*      *     |    :  *
------*---------+----+----*--
-3         |    2    3
|          *

There!

There is a "cusp" at (0,4).
.