What is the equation of the line that is normal to #f(x)= x-e^sqrt(x+4)/(x-4) # at # x= 5 #?

Question

What is the equation of the line that is normal to #f(x)= x-e^sqrt(x+4)/(x-4) # at #  x= 5  #?

Christopher Agresta · Accepted Answer

#y=5−e^3-6/(6+5e^3)*(x-5)# #f(x)=x−(e^(sqrt(x+4)))/(x−4);quadquadquad# #a=5# #y=f(a)-1/(f'(a))*(x-a)# #f'(x)=1-(e^(sqrt(x+4))1/2(x+4)^(-1/2)(x-4)-e^(sqrt(x+4)))/(x-4)^2# #f'(5)=1-(e^(sqrt(5+4))1/2(5+4)^(-1/2)(5-4)-e^(sqrt(5+4)))/(5-4)^2# #f'(5)=1-(e^3(1/2)(9)^(-1/2)-e^3)/1=1-(e^3(1/(2*3))-e^3)# #f'(5)=1-e^3(1/6-1)=1-e^3(-5/6)=1+5/6e^3=(6+5e^3)/6~~17.7379# #f(5)=5−(e^(sqrt(5+4)))/(5−4)=5−e^3~~-15.085# #y=5−e^3-6/(6+5e^3)*(x-5)#

Samantha Adkison · Answer

undefined To find the equation of the line that is normal to the function f(x) at x = 5, we need to determine the slope of the tangent line at x = 5 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 at x = 5, we can take the derivative of the function f(x) with respect to x and evaluate it at x = 5.

After finding the derivative, we substitute x = 5 into the derivative to obtain the slope of the tangent line.

Finally, we take the negative reciprocal of this slope to find the slope of the normal line.

The equation of the line that is normal to f(x) at x = 5 can be written in point-slope form using the point (5, f(5)) and the slope of the normal line.