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Mathematics 19 Online
OpenStudy (anonymous):

The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)/g(x). Find each of the following: u'(1)= v'(1)=

OpenStudy (anonymous):

OpenStudy (anonymous):

ho ho ho have fun actually this is not that hard, lets do one

OpenStudy (anonymous):

you pick it

OpenStudy (anonymous):

number one is 1. however, I can't figure #2

OpenStudy (anonymous):

idea is this \(f'(1)\) is the slope of the line at \(x=1\) it looks like \(f(1)=2\) right?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

and \(g'(1)=-\frac{3}{2}\) if i am reading the graph correctly also \(f(1)=g(1)=2\) for sure

OpenStudy (anonymous):

so \((\frac{f}{g})'=\frac{gf'-fg'}{t^2}\) replace \(f\) and \(g\) by \(2\), \(f'\) by 1 and \(g'\) by \(-\frac{3}{2}\) and do the annoying arithmetic

OpenStudy (anonymous):

typo there, of course it should be \[(\frac{f}{g})'=\frac{gf'-fg'}{g^2}\]

OpenStudy (anonymous):

I got -3/2

OpenStudy (anonymous):

i cant do that arithmetic, but i assume you are correct

OpenStudy (anonymous):

its wrong :/

OpenStudy (anonymous):

actually maybe not \(g'(1)=-\frac{3}{2}\)

OpenStudy (anonymous):

2*1-2*(-3/2)/2^2 right?

OpenStudy (anonymous):

\[\frac{2\times 1+\frac{3}{2}\times 2}{2^2}\]

OpenStudy (anonymous):

5/2

OpenStudy (anonymous):

\[\frac{2+3}{4}=\frac{5}{4}\] is what i get

OpenStudy (anonymous):

nope, thats wrong.

OpenStudy (anonymous):

damn arithmetic is a killer

OpenStudy (anonymous):

:/

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