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CALCULUS

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An equation of the line tangent to the graph of (5x - 1)/(3x + 1) at the point where x = 1 is? Any ideas?

  • CALCULUS -

    sure, I have an idea. Find the derivative to get the slope of the tangent line! Then use the point-slope form of the line's equation:

    y = (5x-1)/(3x+1)
    y(1) = 4/4 = 1

    y' = 8/(3x+1)^2
    y'(1) = 8/16 = 1/2

    tangent line is thus

    (y-1) = 1/2 (x-1)

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