# What are the points of inflection, if any, of f(x)= -14x^3 + 19x^2 - x - 2 ?

May 6, 2016

Points of inflection are $\left(0.64 , 1.47\right)$ and $\left(- 0.21 , - 0.812\right)$

#### Explanation:

To find points of inflection, differentiate the expression to get the gradients, and then find the points where the gradient is zero.
$f \left(x\right) = - 14 {x}^{3} + 19 {x}^{2} - x - 2$

$f ' \left(x\right) = - 42 {x}^{2} + 38 x - 1$

for points of inflection, $- 42 {x}^{2} + 38 x - 1 = 0$

$42 {x}^{2} - 38 x + 1 = 0$

This does not factorise so use the quadratic equation $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{38 \pm \sqrt{1444 - 168}}{84}$

$x \approx \frac{38 \pm 35.72}{84}$

$x \approx 0.64$ or $x \approx - 0.21$

Substitute back into the original expression to find the values of $f \left(x\right)$

Points of inflection are $\left(0.64 , 1.47\right)$ and $\left(- 0.21 , - 0.812\right)$