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

Question

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

Scarlett Allison · Accepted Answer

#2x-y+3=0#. The inserted Socratic graph depicts all the features in the explanation.

To make f real, #x > -2#. At #x =-1, y = 1#. And the foot of the normal is #P(-1, 1)#. #f'=((x+2)^(-1/2))' =-1/2(1/((x+2)sqrt(x+2)))#. So, the slope of the normal is #-1/(f')=2(x+2)sqrt(x+2)=2#, at #P(-1, 1)#. Its equation is #y-1=2(x-(-1))#, giving #2x-y+3=0#.. graph{(ysqrt(x+2)-1)(2x-y+3)=0x^2 [-10, 10, -5, 5]}

Ezekiel Alexander · Answer

The equation is #y = 2x + 3#.

We start by finding the y-coordinate at the given x-point. #f(-1) = sqrt(1/(-1 + 2)) = sqrt(1) = 1# We can start by simplifying the function. #f(x) = 1/sqrt(x + 2)# We now use the chain rule to differentiate #sqrt(x + 2)# before using the quotient rule to differentiate #1/sqrt(x + 2)#. Let #y = u^(-1/2)# and #u = x + 2#. Then #dy/(du) = -1/(2u^(3/2))# and #(du)/dx = 1#. #dy/dx= dy/(du) xx (du)/dx# #dy/dx = 1 xx -1/(2u^(3/2))# #dy/dx = -1/(2(x + 2)^(3/2))# The slope of the tangent is given by evaluating #x= a# inside the derivative. #m_"tangent" = -1/(2(-1 + 2)^(3/2))# #m_"tangent" =- 1/(2(1)^(1/2))# #m_"tangent" =- 1/2# The normal line is perpendicular to the tangent, that's to say its slope is the negative reciprocal of that of the tangent. #m_"normal" = -1/(m_"tangent")# #m_"normal" = -1/(-1/2)# #m_"normal" = 2# The equation of the normal line can be found using a point #(-1, 1)# and the slope #(2)#. #y - y_1 = m(x- x_1)# #y - 1= 2(x - (-1))# #y - 1 = 2(x + 1)# #y - 1= 2x + 2# #y = 2x + 3# Hopefully this helps!

Charles Arocha · Answer

undefined The equation of the line normal to f(x) = sqrt(1/(x+2)) at x = -1 is y = -2(x+1) + 1.