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Calculus

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Find the equation of the line tangent to

f(x) = xe^-x

at the point where x = 0. What does this tell you about the behavior of the graph when x = 0?

  • Calculus - ,

    first of all , when x = 0 , f(0) = 0
    so the point is (0,0)

    f'(x) = x(-e^-x) + e^-x
    when x = 0
    f'(0) = 0 + 1 = 1
    equation of tangent : y = x

    at (0,0) the function is increasing

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