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

Answer 1

Emma Aldridge

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

#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)#

An expression for 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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Answer 2

Isabella Ackerman

The equation of the tangent line of f(x)=sqrt(x+1)-sqrt(x+2) at x=2 is y = -1/6x + 5/6.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.