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the following limit represents the derivative of some function f at some number a lim...

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questionandan... | Student, Undergraduate | (Level 2) Honors

Posted October 6, 2011 at 7:17 AM via web

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the following limit represents the derivative of some function f at some number a lim t->1 [(t^4)+t-2]/t-1. f(x)=? a=?

also for what values of x is the tangent line of the graph of f(x)= (6x^3)-(27x^2)-71x-36. parallel to the line y=x+1.6? enter the x values in order, smallest first, to four places of accuracy:

x1=? x2=?

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beckden | High School Teacher | (Level 1) Educator

Posted October 6, 2011 at 9:57 AM (Answer #1)

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`(df(x))/(dx)=lim_(h->0)(f(x+h)-f(x))/h`

`lim_(t->1)(t^4+t-2)/(t-1)` first substitute t=h+1 to get

`=lim_(h+1->1)((h+1)^4+(h+1)-2)/(h+1-1)`

`=lim_(h->0)((h^4+4h^3+6h^2+4h+1)+h+1-2)/h`

`=lim_(h->0)(h^4+4h^3+6h^2+5h)/h`

Noting that `(1+h)^4+(1+h)=h^4+4h^3+6h^2+4h+1+1+h`

`=h^4+4h^3+6h^2+5h+2` and `(1)^4+(1)=2`

so if `f(1+h)-f(1))=h^4+4h^3+6h^2+5h`

So our function is `f(x)=x^4+x` and `a=1`

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beckden | High School Teacher | (Level 1) Educator

Posted October 6, 2011 at 10:04 AM (Answer #2)

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what values of x is the tangent line of the graph of `f(x)= (6x^3)-(27x^2)-71x-36` . parallel to the line `y=x+1.6` ?

`f'(x)=18x^2-54x-71` Take the derivative.

So when is f'(x)=1 (1 is the slope of the line `y=x+1.6` )

`18x^2-54x-71=1` Find when the derivative is =1

`18x^2-54x-72=0` Write in standard form

Divide everything by 18

` x^2-3x-4=0 `

 

Factor into

`(x-4)(x+1)=0`

So `x_1 = -1` , `x_2 = 4` By the zero product property

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