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OpenStudy (anonymous):
find linearization of (64+3x^2)^(-1/2)
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OpenStudy (anonymous):
this means find the equation of the line tangent to the curve
but to do this you need a point as well
OpenStudy (anonymous):
at a=0 that would be the point right?
OpenStudy (anonymous):
i plug zero into the derivative to get the slope?
OpenStudy (anonymous):
i don't know, you cannot assume the point is a = 0 you must be told
OpenStudy (anonymous):
you can only find the linearization at a point, so you have to be told both the function and a point
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OpenStudy (anonymous):
that is the point i was given i just forgot to write it
OpenStudy (anonymous):
would this be correct.. y=1/8+512(x-0)?
OpenStudy (anonymous):
oh ok
find the derivative, replace x by 0 and get the slope, then use point slope formula
lets see what we get
OpenStudy (anonymous):
\[f(x)=(64+3x^2)^{-1/2}\]
\[f'(x)=-\frac{3x}{(64+3x^2)^{\frac{3}{2}}}\]
OpenStudy (anonymous):
hmm i get
\[f'(0)=0\] did i make a mistake?
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OpenStudy (anonymous):
oh no i got zero too
OpenStudy (anonymous):
so plug that into point slope formula is that what i do?
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