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What is the equation of the line normal to # f(x)=-sqrt((x+1)(x+3)# at # x=0#?

Answer 1

Ezekiel Alexander

#y=sqrt(3)/2x-sqrt(3)#

Given #f(x)=-sqrt((x+1)(x+3))# we get #f(0)=-sqrt(3)#

Now we will compute the slope of the normal line:

#f'(x)=-((x+1)(x+3))^(-1/2)*(x+3+x+1)#so #f'(x)=-(4+2x)/(2sqrt((x+1)(x+3))#

#f'(0)=-2/sqrt(3)# so the slope of our normal line is given by

#m_N=sqrt(3)/2#

#y=sqrt(3)/2x+n#

The Point is given by #(0,-sqrt(3))#

and we get #n=-sqrt(3)#

therefore our line is given by

#y=sqrt(3)/2x-sqrt(3)#

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Answer 2

Isaiah Allen

The equation of the line normal to f(x) = -sqrt((x+1)(x+3)) at x=0 is y = -1.

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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.