`f(x) = 6/(x + 2), (0, 3)` Find an equation of the tangent line to the graph of f at the given point.

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Chapter 2, 2.1 - Problem 32 - Calculus of a Single Variable (10th Edition, Ron Larson).
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hkj1385 | (Level 1) Assistant Educator

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The given function is:-

f(x) = 6/(x+2)

differentiating both sides w.r.t 'x' we get

f'(x) = -6/{(x+2)^2} 

Now, slope of the tangent at the point (0,3) = f'(0) = -6/4 = -3/2

Thus, equation of the tangent at the point (0,3) and having slope = -3/2 is :-

y - 3 = (-3/2)*(x-0)

or, 2y - 6 = -3x

or, 2y + 3x = 6 is the equation of the tangent to the  given curve at (0,3)

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