Hey everyone! My name is Ben, I'm a senior in high school working on an BC Calc problem and am having some general questions about it. The question reads: Let f be a function defined on the closed interval [-3,4] with f(0) = 3. The graph of f', the derivative of f, consists of one line segment and a semi circle, as shown above. Then they provide this image: http://t0.gstatic.com/images?q=tbn:ANd9GcS5nCFcatsgGUIOnZXhKAObqzlF7qkiuW5MOX13BQcePo0e-N3jGQ&t=1 Letter D reads: Find f(-3) and f(4). Show the work that leads to your answers. To find f(-3), I found an equation for the line of f'(x) on x: [-3,0]. I found the equation to be y' = -x - 2. I then recovered the original function on that same interval, and found y = -x2/2 - 2x + C. I used the point (0,3) to find C because of the f(0) = 3 and found that f(x) = -x2 - 2x + 3 (all of this only on the interval [-3,0]. I then plugged in -3 into this recovered equation to find f(-3) = 9/2. For whatever reason though, the second question baffled me. I tried to integrate geometrically again (as from 0 to 4 it's a semicircle of r=2 cut out of a rectangle of dimensions 2x4) and came up with the answer 8-2π (that's supposed to be a pi symbol...). I guess my primary question is, do I add to that the integral of the other side? Would f(4) be the integral from [-3,0] + [0,4], or simply the integral from [0,4]? Do I use the equation of a circle to do something similar; do I have to integrate that? Thanks so much! It's a wonder that a site like this exists :P
First of all, I assume you mean f(0) = 3. How did you get 9/2 when integrating f '(x) from ‒3 to 0 ? From ‒3 to ‒2, there is an area of 1/2 above the x-axis. From ‒2 to 0, there is an area of 2, which is below the x-axis. That translated to an integral of ‒2. That gives a result of ‒3/2. In general: [tex]\int_a^bf'(x)\,dx=f(b)-f(a)[/tex] That's about all you need.
Sorry about the notation mistake, having to hold down shift to do all my parentheses must have thrown me off; a mere typo. Alas, I mixed up what I was doing when typing out my method. To find f(-3), I found an equation for the line of f'(x) on x: [-3,0]. I found the equation to be y' = -x - 2. I then recovered the original function on that same interval, and found y = -x^{2}/2 - 2x + C. I used the point (0,3) to find C because of the f(0) = 3 and found that f(x) = -x^{2} - 2x + 3 (all of this only on the interval [-3,0]. I then plugged in -3 into this recovered equation to find f(-3). Does that make sense? I'm sorry my work is all kind of jumbled on my sheet, I wrote out my process for a different part of the problem but matched it with the answer I got for this one... So in light of having done that, how would I find f(4)? The graph to the right of the origin is a semi circle with radius 2; x^{2} + y^{2} = 4. which would lead to y = sqrt (4 - x^{2}). But that's as far as I know how to go
Finding f(‒3):Yes, that's one way to find f(‒3). What I suggested went along with evaluating the integral graphically. There is more area below the x-axis than above on the interval [‒3, 0]. The area is ‒3/2. Therefore, f(‒3) ‒ f(0) = ‒3/2 . Plug 3 in for f(0) and solve for f(‒3). Did you get f(‒3) = 3/2 ? Find f(4) in a similar way. What is the area between f '(x) and the x-axis on the interval [0, 4] ? It looks to me like it's the area of a 4×2 rectangle minus half the area of a circle of radius 2. (Of course this answer is negative since it's below the x-axis.) Since f(x) is the anti-derivative of f '(x), the area is equal to f(4) ‒ f(0) .