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  • 5 years ago

let f(x)=2x^2-3x +1 f'(x)=lim f(x+h) -f(x)/h h->0

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  1. dumbcow
    • 5 years ago
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    Replace x with (x+h) -> f(x+h) = 2(x+h)^2 - 3(x+h) + 1 Expand -> f(x+h) = 2x^2 + 4xh + 2h^2 - 3x - 3h +1 subtract original function ->(2x^2+4xh+2h^2-3x-3h+1) - (2x^2-3x+1) add like terms **Notice how some terms will cancel ->4xh + 2h^2-3h Factor out an h -> h(4x + 2h - 3) This is important because now we can cancel the h in the denominator Now re-evaluate the lim of (4x + 2h - 3) as h->0 lim = 4x -3 Hope this helps

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