What is the equation of the tangent line to the function 6e^(4x) at x=0?

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  • Como
    Lv 7
    4 weeks ago
    Favorite Answer

    :-

    f ' (x) = 24 e^(4x)

    f ' (0) = 24 = m , the gradient at point (0,6)

    y - 6 = 24 x

    y = 24x + 6

  • alex
    Lv 7
    4 weeks ago

    f(0)=6

    f'(x)=24e^(4x)--->f'(0)=24

    Tangent

    y=24x+6

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