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CALCULUS help please

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Find the following limit where f(x) = 4x^2 − 3x.

limit-->(f(x+ delta x)-f(x))/(delta x)
delta x

**delta x means the triangle symbol with x next to it***

  • CALCULUS help please -

    what is delta x approching? if it is approching 0 this is the definition of the derivative. so it will be 8x-3. I got this by using the rule that the derivative of ax^n=n*a*x^n-1. but they probboly want you to do it without taking the derivative.so lets start by writing it out.
    y=delta*x you don't have to do this I just don't want to keep writing delta*x

    (4(x+y)^2-3(x+y)-(4x^2-3x))/(y)


    next we want to rewrite this so we can take the limit.

    sqrare the x+y
    distribute the 4
    distribute the 3
    get rid of some terms that cancal out and you should get 8*x+4*y-3.

    now all we have to do is take the limmit as y approches 0.

    i get 8x+4*0-3=8*x-3.

    this verifys that the derivative of 4*x^2-3x=8*x-3

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