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October 2nd, 2011, 06:37 PM   #1
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Derivative and table of values

Hey guys, I missed a day of class last week and I can't seem to do this problem in the attached file from the text, I don't really know where to begin. I know it has something to do with first fundamental theorem of calculus. If someone could show me the approach I would really appreciate it, it's been bothering me all weekend.
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 October 2nd, 2011, 07:22 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Derivative and table of values I think I would treat this as a series of initial value problems. For $0\le x<1$ we have: $\frac{df}{dx}=-x$, $f(0)=2$ $\int_{f(0)}\,^{f(1)}\,df=-\int_0\,^1 x\,dx$ $f(1)-f(0)=-\frac{1}{2}$ $f(1)=f(0)-\frac{1}{2}=2-\frac{1}{2}=\frac{3}{2}$ For $1\le x<3$ we have: $\frac{df}{dx}=-1$, $f(1)=\frac{3}{2}$ $\int_{f(1)}\,^{f(3)}=-\int_1\,^3\,dx$ $f(3)-f(1)=-2$ $f(3)=\frac{3}{2}-2=-\frac{1}{2}$ Because the slope is constant here, we may use: $f(2)=\frac{f(1)+f(3)}{2}=\frac{\frac{3}{2}-\frac{1}{2}}{2}=\frac{1}{2}$ For $3\le x<5$ we have: $\frac{df}{dx}=x-4$, $f(3)=-\frac{1}{2}$ $\int\,df=\int x-4\,dx$ $f(x)=\frac{1}{2}x^2-4x+C$ $f(3)=\frac{1}{2}3^2-4(3)+C$ $-\frac{1}{2}=\frac{9}{2}-12+C$ $C=7$ $f(x)=\frac{1}{2}x^2-4x+7$ $f(4)=\frac{1}{2}4^2-4(4)+7=-1$ $f(5)=\frac{1}{2}5^2-4(5)+7=-\frac{1}{2}$ For $5\le x\le 6$ we have: $\frac{df}{dx}=1$, $f(5)=-\frac{1}{2}$ $\int_{f(5)}\,^{f(6)}\,df=\int_5\,^6\,dx$ $f(6)-f(5)=6-5$ $f(6)=-\frac{1}{2}+1=\frac{1}{2}$ Thus, we have: $f(0)=2$ $f(1)=\frac{3}{2}$ $f(2)=\frac{1}{2}$ $f(3)=-\frac{1}{2}$ $f(4)=-1$ $f(5)=-\frac{1}{2}$ $f(6)=\frac{1}{2}$
 October 2nd, 2011, 07:37 PM #3 Senior Member   Joined: Sep 2011 From: New York, NY Posts: 333 Thanks: 0 Re: Derivative and table of values wow.

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