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Posted by on Monday, October 15, 2012 at 11:06am.

Determine an equation of the tangent line to the curve of f (x) = 3e^4x − 3 at the point where the curve crosses the y-axis.

  • Calculus - , Monday, October 15, 2012 at 11:11am

    The curve crosses the y axis when x = 0.
    f(0) = 3*e^0 - 3 = 0 @ x=0
    dy/dx = 12 e^4x = 12 @ x=0

    The straight tangent line at the origin (the tanglent point) is

    y = 12 x.

  • Calculus - , Monday, October 15, 2012 at 11:13am

    y=0 when x=0, so we are looking at the line through (0,0)

    f'(x) = 12e^4x
    f'(0) = 12

    the line is thus y=12x

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