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Calculus

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If the function f is defined by f(x)=x^2-2x, find limit h->0
(f(3+h)-(f(3-h))/2h

  • Calculus - ,

    first, f(3+h)=(3+h)^2 -2(3+h)
    = 9+6h+h^2-6-2h = h^2+4h+3= (h+3)(h+1)
    second f(3-h)=(3-h)^2 -2(3-h)=
    = 9-6h+h^2-6+2h=h^2-4h+3=(h-3)(h-1)

    so first -second then is
    h^2-h^2+4h+4h +3-3=8h

    finally, lim 8h/2h=4

    check all that.

  • Calculus - ,

    Thank you for your help!

    For the second part I thought it would be -(9+6h-h^2)-6+2h?

  • Calculus - ,

    you put a minus sign in front of the first term? Why?

  • Calculus - ,

    and, the 6h term in (3-h)^2=9-6h+h^2

  • Calculus - ,

    (f(3+h)-(f(3-h))/2h

    I put a minus sign because it has a negative in front of the f(3-h) so it's -(f(3-h))

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