What are the points of inflection of #f(x)=(5x-3)/(5x-1)^2

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Question

What are the points of inflection of #f(x)=(5x-3)/(5x-1)^2

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William Abdalla · Accepted Answer

#f(x)=(5x−3)/(5x−1)^2# #f^'(x)=(5x(5x-1)^2-10(5x-3)(5x-1))/(5x−1)^4# #f^'(x)=(5(5x-1)[x(5x-1)-2(5x-3)])/(5x−1)^4# #f^'(x)=(5[5x^2-x-10x+6])/(5x−1)^3# #f^'(x)=(5[5x^2-11x+6])/(5x−1)^3# #f^''(x)=(5[(10x-11)*(5x-1)^3-(5x^2-11x+6)*3(5x-1)^2*5])/(5x−1)^3# #f^''(x)=(5(5x-1)^2[50x^2-55x-10x+11-15(5x^2-11x+6)])/(5x−1)^3# #f^''(x)=(5[50x^2-65x+11-75x^2+165x-90])/(5x−1)# #f^''(x)=(5[-25x^2+100x-79])/(5x−1)# Finding the quadratic function's roots is our next task. #x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)# #x_(1,2)=(-100+-sqrt(100^2-4*(-25)*(-79)))/(2(-25))# #x_(1,2)=(-100+-sqrt(2100))/(2(-25))=(-100+-sqrt(21*100))/(-50)# #x_(1,2)=(10*cancel10+-cancel10sqrt(21))/(5(cancel10))=(10+-sqrt(21))/(5)# #x_1=(10-sqrt(21))/(5)~~1.0834# #x_2=(10+sqrt(21))/(5)~~2.9165# #x_3=1/5=0.2#

Axel Alvarez · Answer

undefined To find the points of inflection, you need to find the second derivative of the function and then solve for where it equals zero.

First derivative: f'(x) = (20x^2 - 16x - 3) / (5x - 1)^3

Second derivative: f''(x) = (200x^3 - 390x^2 + 180x - 16) / (5x - 1)^4

Set the second derivative equal to zero and solve for x to find the points of inflection.