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using four step rule in calculus. use the limit definition to compute y=3x^2-x-2

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user8279914 | eNotes Newbie

Posted July 16, 2013 at 5:09 PM via web

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using four step rule in calculus.

use the limit definition to compute y=3x^2-x-2

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 16, 2013 at 5:27 PM (Answer #1)

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The request of the problem seems to be incomplete because you maybe need to evaluate the derivative of the given function using limit definition.

`f'(x) = lim_(Delta x -> 0) (f(x + Delta x) - f(x))/(Delta x)`

`f(x + Delta x) = 3(x + Delta x)^2 - (x + Delta x) - 2`

`f(x + Delta x) = 3x^2 + 6x*Delta x + 3Delta^2 x - x - Delta x - 2`

Replacing `3x^2 + 6x*Delta x + 3Delta^2 x - x - Delta x - 2 for f(x + Delta x)` yields:

`f'(x) = lim_(Delta x -> 0) (3x^2 + 6x*Delta x + 3Delta^2 x - x - Delta x - 2 - 3x^2 + x + 2)/(Delta x)`

Reducing duplicate members yields:

`f'(x) = lim_(Delta x -> 0) (6x*Delta x + 3Delta^2 x - Delta x)/(Delta x)`

Factoring out `Delta x` yields:

`f'(x) = lim_(Delta x -> 0) Delta x*(6x + Delta x - 1)/(Delta x)`

Reducing duplicate factors yields:

`f'(x) = lim_(Delta x -> 0) (6x + Delta x - 1)`

Replacing 0 for `Delta x` yields:

`f'(x) = 6x + 0 - 1 => f'(x) = 6x - 1`

Hence, evaluating the derivative of the given function, using limit definition, yields `f'(x) = 6x - 1.`

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