# What are the points of inflection, if any, of f(x)=4x^3 + 21x^2 - 294x +7 ?

Dec 7, 2017

$\left(\frac{7}{2} , - \frac{2373}{4}\right) \mathmr{and} \left(- 7 , 1722\right)$ are points of minimum and maximum respectively, though none are at a point where inflection occurs, but the graph continues going in the same general direction.

#### Explanation:

An inflection occurs when $f ' \left(x\right) = 0$

To determine this, we must differentiate $f \left(x\right) \mathrm{dx}$

$f \left(x\right) = 4 {x}^{3} + 21 {x}^{2} - 294 x + 7$

$f ' \left(x\right) = 12 {x}^{2} + 42 x - 294$

$12 {x}^{2} + 42 x - 294 = 0$

$6 {x}^{2} + 21 x - 147 = 0$

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

We get:
$x = \frac{- 21 \pm \sqrt{{21}^{2} - 4 \left(6 \cdot - 147\right)}}{2 \left(6\right)}$

$x = \frac{- 21 \pm \sqrt{441 - 4 \left(- 882\right)}}{12}$

$x = \frac{- 21 \pm \sqrt{441 + 3528}}{12}$

$x = \frac{- 21 \pm \sqrt{3969}}{12}$

$x = \frac{- 21 + 63}{12} \mathmr{and} \frac{- 21 - 63}{12}$

$x = \frac{42}{12} \mathmr{and} - \frac{84}{12}$

$x = \frac{7}{2} \mathmr{and} - 7$

$f \left(\frac{7}{2}\right) = 4 {\left(\frac{7}{2}\right)}^{3} + 21 {\left(\frac{7}{2}\right)}^{2} - 294 \left(\frac{7}{2}\right) + 7 = - \frac{2373}{4}$

$f \left(- 7\right) = 4 {\left(- 7\right)}^{3} + 21 {\left(- 7\right)}^{2} - 294 \left(- 7\right) + 7 = 1722$

The points of inflection are:
$\left(\frac{7}{2} , - \frac{2373}{4}\right) \mathmr{and} \left(- 7 , 1722\right)$

However, this only tells us if a point is either a maximum, minimum or inflection.

The second derivative can help us with this. If $f ' ' \left(x\right) = 0$ it is inflection, $f ' ' \left(x\right) > 0$ is a minimum, $f ' ' \left(x\right) < 0$ is a maximum.

$f ' ' \left(x\right) = 24 x + 42$

$f ' ' \left(\frac{7}{2}\right) = 24 \left(\frac{7}{2}\right) + 42 = 126$ (Therefore a minimum)

$f ' ' \left(- 7\right) = 24 \left(- 7\right) + 42 = - 126$ (Therefore a maximum)

There no points of inflection as such, only minimum and maximum points.