determine the equation of the tangent line to the curve defined by f(x)=4x+8 sqrt(x) at the point where x=25 and f(x)=140
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f(x)=4x+8√x; f'(x)=4+4/√x; f'(25)=4+4/5=4.8.

This is the slope at x=25, so the tangent line has the general equation g(x)=4.8x+a where a is a constant.

But g(x) has to pass through (25,140): 140=4.8*25+a; a=140-120=20.

The equation of the tangent line is g(x)=4.8x+20.

by Top Rated User (764k points)

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