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# The S&Pindex for a co. in recent years is...

jjmgingrich | Student, Undergraduate | Salutatorian

Posted May 11, 2013 at 2:41 AM via web

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The S&Pindex for a co. in recent years is (2007:SPIn(80))(2008:SPIn(40))(2009:SPIn(20))(2010:SPIn(30)). The index is approximated by the function f(x) = 13x^2 - 82x + 150, where x stands for years since 2006.  QUESTION:  Find f' (x) using the derinition of the derivative.; Evaluate the derivative at x=1 and interpret the result; find the rate of change of the index in 2010... any qestions, PLEASE MESSAGE ME or text me, 717-385-0641

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### 1 Answer |

pramodpandey | College Teacher | Valedictorian

Posted May 11, 2013 at 4:11 AM (Answer #1)

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Def of derivative.

`lim_{h->0}{f(x+h)-f(x)}/h=f'(x)`

`` We have given function

f(x) = 13x^2 - 82x + 150,

Thus

`f'(x)=lim_{h->0}{13(x+h)^2-82(x+h)+150-(13x^2-82x+150)}/h`

`=lim_{h->0}{26hx+169h^2-82h}/h`

`=lim_{h->0}{(h(26x-82))/h}`

`=26x-82`

Thus rate of change at x=1

`f'(x)}_{x=1}=26-82`

`=-56`

Index function is a decreasing function at x=1.

Index in 2010 is

`f(4)=13(4)^2-82xx4+150`

`=30`

rate of change in 2010

`f'(x)}_{x=4}=26xx4-82`

`=22`

rate of change in 2010 is positive ,index is now an increasing function.

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