equation to tangent line

Question

Find the slope and an equation of the tangent line to the graph of the function f at the specified point.
f(x) = √x + (1/√x); (25,26/5)

 

slope=
equation y =

Answer 1

f(x) = √x + (1/√x). 

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

k=f'(x₀)=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=k

Tangent  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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