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

Question

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

William Abdalla · Accepted Answer

#y=\sqrt6(x-2)#

The given function

#f(x)=\frac{2-x}{\sqrt{2x+2}}#

setting #x=2# in the given function, we get #y#-coordinate of point

#y=f(2)#

#=\frac{2-2}{\sqrt{2\cdot 2+2}}=0#

Now, differentiating above equation w.r.t. #x# we get slope #dy/dx# of tangent as follows

#dy/dx=d/dxf(x)#

#=\frac{\sqrt{2x+2}(-1)-(2-x)\frac{1}{\sqrt{2x+2}}}{(\sqrt{2x+2})^2}#

#=\frac{-(2x+2)-(2-x)}{(2x+2)^{3/2}}#

#=\frac{-x-4}{(2x+2)^{3/2}}#

setting #x=2# in above equation, we get slope of tangent at #(2, 0)#

#=\frac{-2-4}{(2\cdot 2+2)^{3/2}}#

#=-1/\sqrt6#

hence the slope #m# of normal at the same point

#=-1/(-1/\sqrt6)=\sqrt6#

hence the equation of normal at #(x_1, y_1)\equiv (2, 0) # & having slope #m=\sqrt6# is given by following formula

#y-y_1=m(x-x_1)#

#y-0=\sqrt6(x-2)#

#y=\sqrt6(x-2)#

Samantha Adkison · Answer

undefined To find the equation of the line that is normal to the function f(x) = (2-x)/sqrt(2x+2) at x=2, we need to determine the slope of the tangent line at x=2 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we can take the derivative of f(x) with respect to x and evaluate it at x=2.

The derivative of f(x) = (2-x)/sqrt(2x+2) is given by:

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

Evaluating f'(x) at x=2:

f'(2) = (2(sqrt(2(2)+2)) + (2-2)(1/2)(2(2)+2)^(-1/2)) / (2(2)+2)
       = (2(sqrt(6)) + 0) / 6
       = (2sqrt(6)) / 6
       = sqrt(6) / 3

The slope of the tangent line at x=2 is sqrt(6) / 3.

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line:

Slope of the normal line = -1 / (sqrt(6) / 3)
                        = -3 / sqrt(6)

Now that we have the slope of the normal line, we can use the point-slope form of a line to find the equation. Since the line is normal to f(x) at x=2, we know that it passes through the point (2, f(2)).

Substituting x=2 into f(x), we get:

f(2) = (2-2)/sqrt(2(2)+2)
      = 0

Therefore, the point of intersection is (2, 0).

Using the point-slope form of a line, we have:

y - y1 = m(x - x1)

Substituting the values, we get:

y - 0 = (-3 / sqrt(6))(x - 2)

Simplifying, we obtain the equation of the line that is normal to f(x) at x=2:

y = (-3 / sqrt(6))(x - 2)