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anonymous

  • 5 years ago

1. find the slope of the tangent to the curve y=1/sqroot(x) at x=a. 2. find the equation of the line at (4,1/2)

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  1. anonymous
    • 5 years ago
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    The slope of the tangent to a curve f(x) at x = a is just f’(a), ie. the derivative evaluated at x = a. Once you know the slope of the tangent line, and you want to find the equation of the tangent line itself, you’d probably want to use the point-slope form y – y1 = m(x – x1) since you know that your tangent line passes through a point (x1, y1). That automatically gives you the equation of the line. Here y = 1/x^(1/2) = x^(-1/2). So y’ = (-1/2) x^(-3/2). Then y’(a) = (-1/2) a^(-3/2) will be the slope of the tangent line at x = a Then the equation of the tangent line is y – 1/2 = (-1/2) * (4)^(-3/2) (x – 4). Notice that we plugged in the derivative at x = 4 for the slope m. Simplify to get the final answer.

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