Thread: sketching graph of a continuous function

Sketch a possible graph of continuous function f that has domain [-3,3] where f(-1)=1

I can't think of what the f(-1)=1 means, can someone help me out?

The original problem includes a picture of the graph of the derivative, but i do not know how to get the picture on the thread. I can describe it though.
y = f'(x)
A negative slant starts at (-3, 2.5) and goes down to (1,-2.5). There is an open circle at the end of the line at (1,-2.5) indicating discontinuity. The end of a parabola starts with another open circle at (1,2.5) goes down to (2,0) and back up to (3,2.5).

Any help is appreciated, I apologize for not posting a picture of the derivative graph but i do not know how.

Originally Posted by Murphie

Sketch a possible graph of continuous function f that has domain [-3,3] where f(-1)=1

I can't think of what the f(-1)=1 means, can someone help me out?

The original problem includes a picture of the graph of the derivative, but i do not know how to get the picture on the thread. I can describe it though.
y = f'(x)
A negative slant starts at (-3, 2.5) and goes down to (1,-2.5). There is an open circle at the end of the line at (1,-2.5) indicating discontinuity. The end of a parabola starts with another open circle at (1,2.5) goes down to (2,0) and back up to (3,2.5).

Any help is appreciated, I apologize for not posting a picture of the derivative graph but i do not know how.

f(-1)=1 means the graph of y = f(x) passes through the point (-1, 1).