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)
Now we have to find roots of quadratic function:
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