# curve sketching

• Jun 7th 2010, 08:25 PM
sydewayzlocc
curve sketching
I am having difficulty with the following :

Sketch the graph, identifying any relative or absolute extrema and points of inflection, of a continuous function f(x) on the interval [0,8] which satisfies these conditions:

f(0)=5, f(2)=2, f(4)=4, f(6)=1, f(8)=3

f'(x)<0 on (0,2)U(4,6) and f'(x)>0 on (2,4)U(6,8)
f'(2) does not exist and f"(x)<0 on (0,2)U(2,5) and f"(x)>0 on (5,8)

im not even sure how to begin?
• Jun 8th 2010, 01:32 AM
p0oint
Ok, its not that hard, because you have so many graphs that match your conditions.

How will begin by putting (0,5), (2,2), (4,4), (6,1) and (8,3) on the coordinate system.

Then you notice something. f'(2) does not exist but f(2)=2. This tells you that there is either a cusp or vertical tangent. What tells you that there is cusp is actually f'(x) changing sign from negative to positive in front of x=2. This will tell you that x=2, y=2 is local extrema (concretely minimum because of the reason above).

You will notice that f'(x) is changing sign also in x=4, local maxima.

x=6, local minima same reasons.

Infletion point: x=5 because f''(x) is changing sign.

You got so many things given that its very easy to draw the graphic.