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 July 21st, 2016, 08:24 AM #1 Newbie   Joined: Jul 2016 From: pickering Posts: 7 Thanks: 2 instantaneous rate of changes Given the function g(x)=x^4-2x^3+1, a)Estimate the instantaneous rate of change at x=1 , b. Will the instantaneous rate of change be the same at any x-value on the function? Explain why or why not. i need help thank you Thanks from manus July 21st, 2016, 08:29 AM #2 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 First tell us what you understand 'instantaneous rate of change ' to mean. July 21st, 2016, 08:29 AM #3 Senior Member   Joined: Apr 2014 From: UK Posts: 967 Thanks: 344 Are you familiar with differentiation? July 21st, 2016, 08:38 AM   #4
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 Originally Posted by weirddave Are you familiar with differentiation?
what do you mean? July 21st, 2016, 08:41 AM   #5
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 Originally Posted by studiot First tell us what you understand 'instantaneous rate of change ' to mean.

I don't really understand this but I know the formula and I can't firgue out how to find it July 21st, 2016, 08:54 AM   #6
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 Originally Posted by ryan1108109 I don't really understand this but I know the formula and I can't firgue out how to find it
what formula? July 21st, 2016, 08:57 AM   #7
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 Originally Posted by skeeter what formula?
this formula : f(x+h)-f(x) / h July 21st, 2016, 09:04 AM #8 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 That would do although you should not blindly apply any formula, without knowing what it means. So is f(x) clear enough? What do you think h might be? remember the question said estimate the rate of change, not calculate its exact value. Thanks from manus July 21st, 2016, 09:13 AM #9 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 If in doubt, ask. September 16th, 2016, 12:14 AM #10 Newbie   Joined: Jun 2016 From: india Posts: 24 Thanks: 4 Given function : $\displaystyle g(x)=x^4-2x^3+1$ To find rate of change, we have to differentiate given function. $\displaystyle \dfrac{d}{dx}\;\{g(x)\}=\dfrac{d}{dx}\;(x^4-2x^3+1)$ $\displaystyle \Rightarrow\;g'(x)=4x^3-6x^2+0$ To find instantaneous rate of change at $\displaystyle x=1$, put $\displaystyle x=1$ in $\displaystyle g'(x)$. So instantaneous rate of change at $\displaystyle x=1$ $\displaystyle =4(1)^3-6(1)^2$ $\displaystyle =4-6$ $\displaystyle =-2$ If you want to study more about rate of change and approximation, this will help you. Rate of Change Formula, Rate of Change of Volume and Surface, Velocity - Actucation Tags instantaneous, rate Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post rrosa522 Pre-Calculus 2 February 23rd, 2016 05:54 AM ioabs Calculus 1 September 7th, 2015 04:08 PM Mr Davis 97 Calculus 2 May 5th, 2014 08:03 PM Pumpkin99 Calculus 1 September 26th, 2011 06:06 PM mathman2 Calculus 13 October 20th, 2009 12:06 PM

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