- #1
devilazy
- 15
- 0
So the question asks me to use the following information to graph the function f over the closed interval [-2, 5]
i) The graph of f is made of closed line segments joined end to end.
ii) The graph starts at the point (-2, 3)
ii) The derivative of f is the step function in the figure shown here.
That's all give as well as a graph, but I don't know how to put a graph on so I'll just state it's points.
x: [-2 to 0) y' = -2
x: (0 to 1) y' = 0
x: (1 to 3) y' = 1
x: (3 to 5] y' = -1
I am totally clueless of what to do, I missed that class and wasn't able to catch up to what I missed. I tried reading through the text to see if it says anything about this but found nothing. I know y' is basically the slope of f(x), and I know how to do integration (but I am not allowed to use it, as we haven't gotten there yet). Anyways, my point is I tried integrating the y' but yea, it just makes no sense to me. I hate graphs.
I would love any idea, I just wish to understand how to do this and then I'll be fine figuring out the rest.
i) The graph of f is made of closed line segments joined end to end.
ii) The graph starts at the point (-2, 3)
ii) The derivative of f is the step function in the figure shown here.
That's all give as well as a graph, but I don't know how to put a graph on so I'll just state it's points.
x: [-2 to 0) y' = -2
x: (0 to 1) y' = 0
x: (1 to 3) y' = 1
x: (3 to 5] y' = -1
I am totally clueless of what to do, I missed that class and wasn't able to catch up to what I missed. I tried reading through the text to see if it says anything about this but found nothing. I know y' is basically the slope of f(x), and I know how to do integration (but I am not allowed to use it, as we haven't gotten there yet). Anyways, my point is I tried integrating the y' but yea, it just makes no sense to me. I hate graphs.
I would love any idea, I just wish to understand how to do this and then I'll be fine figuring out the rest.