# Homework Help: Derivative/Antiderivative Graphs

1. May 14, 2007

### t_n_p

1. The problem statement, all variables and given/known data

http://img149.imageshack.us/img149/1751/untitled3zt8.jpg [Broken]

If g(x) is an antiderivative of f(x) and g(0)=2, detemine the x-coordinates of the points on the graph of g(x) corresponding to its local max and min values.
3. The attempt at a solution

I understand that stationary points occur when y=0 on the f(x) graph (i.e. at x=1 and x=3.5). I also know that at x=1 it's a maximum and at x=3.5 it's a minimum, but how do I find the corresponding y-values?

Last edited by a moderator: May 2, 2017
2. May 14, 2007

### HallsofIvy

By "integrating". You could, of course, find the equation for f itself since it is "piecewise linear" but that isn't necessary. The integral of f is the area under the graph and, from x= 0 to x= 1, that is just a the area of a triangle. Don't forget to add the initial value, g(0)= 2. One you know g(1), you can find g(3.5) by "adding" the area under the curve from 1 to 3.5. I put "adding" in quotes because that value is actually negative, since the curve is below the x-axis.

To find g(1), Find the area of the triangle with base 1 and height 1 and add that to g(0)= 2. To find g(3.5), find the area of the trapezoid with height 1, bases 2.5 and 1, and subtract that from g(1).

3. May 14, 2007

### t_n_p

ah gotcha thanks!

A turning point on the graph of f refers to a point of inflexion on the graph g, so does that mean that g(x) is increasing for (0,1), (3.5,4) and (4,6)?

Also, how would I find where g(x) is concave up/down?

4. May 14, 2007

### Hootenanny

Staff Emeritus
Well, what does the second derivative of g(x) tell you?

5. May 14, 2007

### t_n_p

Well I do know f"(x)>0 represents concave up and f"(x)<0 represents concave down, but how can I get f"(x) (i.e. the derivative of g)

P.S. are my intervals for increasing g(x) correct?

6. May 14, 2007

### Office_Shredder

Staff Emeritus
But...

Couldn't you just save yourself a lot of time and just answer the question instead?

7. May 14, 2007

### t_n_p

Yeah I know, just wanted to further my knowledge

8. May 14, 2007

### Hootenanny

Staff Emeritus
Yes, your intervals look good to me. You don't need to get f''(x), you only need f'(x). Since g''(x) = f'(x), he gradient of f(x) on your plot represents the second derivative of g(x). Does that make sense?

Last edited: May 14, 2007
9. May 14, 2007

### t_n_p

So g(x) is concave up for positive gradients of f(x), that is (3,4) and concave down for negative gradients of f(x), that is (1,2) and (4,6).

10. May 14, 2007

### Hootenanny

Staff Emeritus
Looks good to me

11. May 14, 2007

### t_n_p

Finally, points of inflexion on the graph of g occur when there are turning points on the graph of f, that is when x=4.

I now need to use all this info and the fundamental theorem of calculus to find values of g(x) at appropriate points to draw g(x). I suppose this is where the y-coordinates come in handy! I'll be back later with my attempted graph!

12. May 14, 2007

### t_n_p

I've got a rough idea of what it looks like already, but with the graph being concave down for (1,2) and concave up for (3,4), what's happening between the interval of (2,3)?

Also how would I go about finding the y value for x=4?
will it be the y value at x=3.5 plus the area of the small triangle, i.e.
0.75+(0.5X0.5X1)?

13. May 14, 2007

### Hootenanny

Staff Emeritus
Well, the second derivative test is inconclusive. However, you can say that g'(x)=-1 for {x : 2 < x < 3}, therefore the slope at this point is constant and the graph is a straight line y = -x.
I haven't checked your numbers, but you method is correct

14. May 15, 2007

### t_n_p

http://img379.imageshack.us/img379/6467/asdfra4.jpg [Broken]

think i got it right

Last edited by a moderator: May 2, 2017