Mathematics
8 Online
OpenStudy (anonymous):
Find the equation of the tangent line to the curve y=(sqrt(x))/(7x+7) at the point (1, f(1)).
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Parth (parthkohli):
Find the derivative of the function. The result will be a function that tells you the slope of the tangent at a given \(x\).
Parth (parthkohli):
Do you need help?
OpenStudy (anonymous):
i get 1/14 as the slope. that's with using the x=1 as the f'(x)
OpenStudy (anonymous):
so does the "point" change from (1,f(1)), to (1,1/14)?
Parth (parthkohli):
Hmm, no -- the derivative is the function that tells you the slope of the tangent at any point, so you have to find f'(1). What you got was f(1).
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OpenStudy (anonymous):
oh, in that case than I get 0
Parth (parthkohli):
That is correct!
OpenStudy (anonymous):
so the slope is 0
Parth (parthkohli):
Exactly!
OpenStudy (anonymous):
but how do I write a line equation without a slope
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OpenStudy (anonymous):
its y=mx+b or y-y1=m(x-x1)
Parth (parthkohli):
Oh, we had to find the equation of the tangent.
Parth (parthkohli):
You know the slope, and a point of the line.
Parth (parthkohli):
That is, (1,0)
OpenStudy (anonymous):
yeah
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Parth (parthkohli):
So use y - y1 = m(x - x1)
OpenStudy (anonymous):
y-0=0(x-1)?
Parth (parthkohli):
Yup! Or y = 0.
OpenStudy (anonymous):
so what would the equation of the line be?
Parth (parthkohli):
y = 0.
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OpenStudy (anonymous):
my math program is telling me y=0 is incorrect...
Parth (parthkohli):
Hmm... I wonder why.
Parth (parthkohli):
Whoops!
OpenStudy (anonymous):
?
Parth (parthkohli):
The point is not (1,0) but (1,1/14). Sorry.
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OpenStudy (anonymous):
so would the equation then look like this? y-1/14=0(x-1)
OpenStudy (anonymous):
then y-1/14=0
then y=1/14
Parth (parthkohli):
Indeed!
OpenStudy (anonymous):
thank you