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February 20th, 2011, 09:18 PM   #1
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Derivatives: Linear Approximation

The demand function for a product is given by

p=f(q)=60??q
where p is the price per unit in dollars for q units. Use the linear approximation to approximate the the price when 899 units are demanded.

Solution: We want to approximate f(899). From
f(q)? L(q)=f(a)+f'(a)(q?a)
and the fact that f'(a)= either -1/sqrt(a) OR 1/(2sqrt(a)) OR -1/(2sqrt(a)) OR 1/sqrt(a)
we choose a= _______.

From f(900)= _____ and f'(900)= ______ we get f(899)? ________ .
Hence, the price per unit when 899 units are demanded is approximately $______
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February 20th, 2011, 09:31 PM   #2
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Re: Derivatives: Linear Approximation

f(q)=60??q

Do you know how to differentiate this? Write it as and have a go.
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