What is the equation of the tangent line of #f(x)=sqrt(x+1)-sqrt(x+2) # at #x=2#?

1 Answer
Lucy
Apr 4, 2018

#4ysqrt3=x(2-sqrt3)+8-6sqrt3#

Explanation:

#f(x)=sqrt(x+1)-sqrt(x+2)#
#f'(x)=1/(2sqrt(x+1))-1/(2sqrt(x+2)#
#f'(x)=(1/2)(sqrt(x+2)-sqrt(x+1))/(sqrt(x+1)sqrt(x+2))#

At #(2,sqrt3-2)#

Sub x=2 into #f'(x)#

#f'(2)=(1/2)(sqrt(2+2)-sqrt(2+1))/(sqrt(2+1)sqrt(2+2))#
#f'(2)=(2-sqrt3)/(4sqrt3)#

Equation of the tangent:

#(y-sqrt3+2)=(2-sqrt3)/(4sqrt3)(x-2)#
#4ysqrt3-12+8sqrt3=2x-4-xsqrt3+2sqrt3#
#4ysqrt3=x(2-sqrt3)+8-6sqrt3#

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