A Angela123 Junior Member Joined Oct 9, 2008 Messages 54 Jul 14, 2009 #1 Use [Delta]y=f'(x)[Delta]x to find a decimal approximation of the radical expression (730)^(1/6) square root of 729=27 1/((1/6)x^(-5/6)) + 27 so I got 26.8611 I don't feel like I did this right.
Use [Delta]y=f'(x)[Delta]x to find a decimal approximation of the radical expression (730)^(1/6) square root of 729=27 1/((1/6)x^(-5/6)) + 27 so I got 26.8611 I don't feel like I did this right.
P PAULK Junior Member Joined Dec 13, 2007 Messages 124 Jul 14, 2009 #2 Angela123 said: Use [Delta]y=f'(x)[Delta]x to find a decimal approximation of the radical expression (730)^(1/6) square root of 729=27 1/((1/6)x^(-5/6)) + 27 so I got 26.8611 I don't feel like I did this right. Click to expand... Neither do I. Start by using: f(x) = x^(1/6) x0 = 729 (yes, good idea) dx = 1 Now f(730) = f(729) + 1 * f'(729) However, f(729) = 729^(1/6) = 3, not 27. So look for an answer slightly bigger than 3.
Angela123 said: Use [Delta]y=f'(x)[Delta]x to find a decimal approximation of the radical expression (730)^(1/6) square root of 729=27 1/((1/6)x^(-5/6)) + 27 so I got 26.8611 I don't feel like I did this right. Click to expand... Neither do I. Start by using: f(x) = x^(1/6) x0 = 729 (yes, good idea) dx = 1 Now f(730) = f(729) + 1 * f'(729) However, f(729) = 729^(1/6) = 3, not 27. So look for an answer slightly bigger than 3.
