Bobosharif S. answered • 02/11/18

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PhD in Math, MS's in Calulus

f(x) = √x + (1/√x).

Equation of the tangent is of the form y=kx+b.

k=f'(x

_{0})=f'(25).f'(x)=1/(2√x)-(1/2)x

^{-(3/2)}f'(25)=1/(2*5)-(1/2)5

^{-3}=1/10-1/250=24/250=12/125=kTangent at the point (25, 26/5):

26/5=(12/125)*25+b.

b=(26-12)/5=14/5.

So equation of the tangent at the point (25, 26/5) is

y=(12/125)x+14/5

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