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 rk183 October 2nd, 2011 06:37 PM

Derivative and table of values

1 Attachment(s)
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.
Thanks.

 MarkFL October 2nd, 2011 07:22 PM

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:

$f(0)=2$

$f(1)=\frac{3}{2}$

$f(2)=\frac{1}{2}$

$f(3)=-\frac{1}{2}$

$f(4)=-1$

$f(5)=-\frac{1}{2}$

$f(6)=\frac{1}{2}$

 aaron-math October 2nd, 2011 07:37 PM

Re: Derivative and table of values

wow.

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