My Math Forum Derivatives: Linear Approximation

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 February 20th, 2011, 09:18 PM #1 Newbie   Joined: Feb 2011 Posts: 6 Thanks: 0 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 \$______
 February 20th, 2011, 09:31 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Derivatives: Linear Approximation f(q)=60??q Do you know how to differentiate this? Write it as $60 - q^{1/2}$ and have a go.

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