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Find an equation for the line tangent to the graph of f(x)=sqrt(x)/5x+5 at the point (3,f(3))
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find \(f'(3)\) that is your slope then your point is \((3,f(3))\) point slope formula will give you the equation for the line
it will be nice and ugly since \(f(3)=\frac{\sqrt{3}}{20}\) if i am reading your function correctly
-1/80sqrt(3)?
is it \(f(x)=\frac{\sqrt{x}}{5x+5}\) ?
yes.. i got that ^ for f'(3)
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(1-3)/(10 sqrt(3) (1+3)^2)
yeah i guess that is right http://www.wolframalpha.com/input/?i=+%281-3%29%2F%2810+sqrt%283%29+%283%2B1%29^2%29
that's the slope?
yup ugly huh?
sqrt(3)/20=(-1)/(80sqrt(3))(3)+b?
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got it! thanks!
you need an \(x\) and a \(y\) in your answer \[y-f(3)=f'(3)(x-3)\]
yw
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