draw curve

(Doh)Sketch the grah of a continuous function f that satisfies all of the stated conditions.

f(0) =2;f(-2) = f(2)=0; f^1(-2) = f^1(0)=f^l(2)=0;
f^l(x)> 0 if -2 <x<0;f^1(x)<0 if x<-2 or x>0;
f^ll(x) > 0if x<-1 or 1 <x<2;
f^ll(x) < 0 if -1<x<1 or x>2

Okay this problem has got me in a wirl wind of fire. Step-by-step help please.

i don't understand how to start the problem but do you start with derivatives

Hello, uniquereason81!

Quote:

Sketch the grah of a continuous function $f(x)$
that satisfies all of the stated conditions.

$f(0) \,=\,2;\;\;f(\text{-}2) \,=\, f(2)\,=\,0\;\;{\color{blue}[1]}$

$f'(\text{-}2) \,= \,f'(0)\,=\,f'(2)\:=\:0\;\;{\color{blue}[2]}$

$f'(x) \;=\;\bigg\{\begin{array}{cccc}> 0 & \text{if }\text{-}2 < x <0 & {\color{blue}[3]}\\< 0 & \text{if }x<\text{-}2\:\cup\: x>0 & {\color{blue}[4]}\end{array}$

$f''(x) \;=\; \bigg\{\begin{array}{cccc}> 0 & \text{if }x < \text{-}1 \:\cup\:1 < x < 2 & {\color{blue}[5]}\\ < 0 & \text{if }\text{-}1 < x < 1 \:\cup \:x > 2 & {\color{blue}[6]}\end{array}$

[1] gives us three points: . $(-2,0),\:(0,2),\:(2,0)$

[2] tells that the tangents are horizontal at those three points

Code:

2| =o= | | | - =o= - - - + - - - =o= - - 2 | 2 |

[3] tells us that the curve is uphill on [-2.0]
[4] says it is downhill everywhere else.

Code:

| * / | \ \ / | \ \ / | \ - -*- - - - + - - - -*- - -2 | 2 \ | \ |

[5] tells us the curve is concave up to the left of -1, and on [1,2]

Code:

| * * / | \ * / | * * * | * - - -*- -+- - + - -+- -*- - -2 -1 | 1 2\ | \ ||

[6] says the curve is concave down everywhere else.

Code:

| * * * | * * * | * * * | * - - -*- -+- - + - -+- -*- - - - 2 -1 | 1 2 * | * | * |

There!

wow thanks alot