My Math Forum instantaneous rate of changes

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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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Quote:
 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

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