SOLUTION: If f(x)=4x+(3/x), find f'(-2), using the definition of derivative. f'(-2) is the limit as x tends to -2. Find the expression and the value of the limit. Use this to find the equati

Question 1094021: If f(x)=4x+(3/x), find f'(-2), using the definition of derivative. f'(-2) is the limit as x tends to -2. Find the expression and the value of the limit. Use this to find the equation of the tangent line to the graph of y=4x+(3/x) at the point (-2,-9.5).
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
From the definition of the derivative, df/dx = lim(h->0) (f(x+h)-f(x))/h
Thus df/dx = 1/h*(4(x+h) + 3/(x+h) - 4x - 3/x)
4(x+h)/h + 3/(h(x+h)) - 4x/h - 3/xh -> 4 + 3/(h(x+h)) - 3/(xh)
Simplify using the common denominator xh^2(x+h):
(4 + 3xh - 3h(x+h))/(xh^2(x+h)) -> 4 - 3h^2/(xh^2(x+h))
The h^2 term cancels and we are left with:
4 - 3/(x(x+h)) The limit as h->0 is 4 - 3/x^2
Thus f'(-2) = 4 - 3/(-2)^2 = 3 1/4
The slope of the tangent line is the value of the derivative at the given point.
Thus m = f'(-2) = 3.25
The equation of the line is y + 9.5 = 3.25(x + 2) -> y = 3.25x - 3
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