# What are the points of inflection, if any, of f(x) = x^6 + 3x^5 - (15/2)x^4 - 40x^3 - 60x^2 + 8x + 5 ?

Nov 25, 2017

Points of inflection occur at $x = - 1 , - 2$ and $2$.

#### Explanation:

At point of inflection $\frac{{d}^{2} f \left(x\right)}{{\mathrm{dx}}^{2}} = 0$

As $f \left(x\right) = {x}^{6} + 3 {x}^{5} - \left(\frac{15}{2}\right) {x}^{4} - 40 {x}^{3} - 60 {x}^{2} + 8 x + 5$

$\frac{\mathrm{df}}{\mathrm{dx}} = 6 {x}^{5} + 15 {x}^{4} - 30 {x}^{3} - 120 {x}^{2} - 120 x + 8$

and $\frac{{d}^{2} f \left(x\right)}{{\mathrm{dx}}^{2}} = 30 {x}^{4} + 60 {x}^{3} - 90 {x}^{2} - 240 x - 120$

Hence points of inflection are given by $30 {x}^{4} + 60 {x}^{3} - 90 {x}^{2} - 240 x - 120 = 0$

or ${x}^{4} + 2 {x}^{3} - 3 {x}^{2} - 8 x - 4 = 0$

Using factor theorem we have zeros at $- 1 , - 2$ and $2$

$\left(x - 2\right) \left(x + 2\right) {\left(x + 1\right)}^{2} = 0$

Hence, points of inflection occur at $x = - 1 , - 2$ and $2$.

Graph not drrawn to scale (shrunk vertically).
graph{x^6+3x^5-(15/2)x^4-40x^3-60x^2+8x+5 [-4, 4, -200, 200]}