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

Dec 14, 2017

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:

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

${x}_{1 , 2} = \frac{- 100 \pm \sqrt{{100}^{2} - 4 \cdot \left(- 25\right) \cdot \left(- 79\right)}}{2 \left(- 25\right)}$

${x}_{1 , 2} = \frac{- 100 \pm \sqrt{2100}}{2 \left(- 25\right)} = \frac{- 100 \pm \sqrt{21 \cdot 100}}{- 50}$

${x}_{1 , 2} = \frac{10 \cdot \cancel{10} \pm \cancel{10} \sqrt{21}}{5 \left(\cancel{10}\right)} = \frac{10 \pm \sqrt{21}}{5}$

${x}_{1} = \frac{10 - \sqrt{21}}{5} \approx 1.0834$

${x}_{2} = \frac{10 + \sqrt{21}}{5} \approx 2.9165$

${x}_{3} = \frac{1}{5} = 0.2$