# How do you find local min and max, point of inflection and concavity for the equation f(x) = 4x^3 + 21x^2 - 294 +7?

Sep 13, 2015

The function has a minimum at $x = 0$
The function has a maximum at $x = \frac{- 7}{2}$
At $x = \frac{- 42}{24}$ there is a point of inflection

#### Explanation:

Given -
$y = 4 {x}^{3} + 21 {x}^{2} - 294 + 7$
Find the first derivative
$\frac{\mathrm{dy}}{\mathrm{dx}} = 12 {x}^{2} + 42 x$
Set it equal to zero. you are doing it to know for which value of 'x' the slope of the curve of the function becomes zero
$\frac{\mathrm{dy}}{\mathrm{dx}} = 6 x \left(2 x + 7\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies 6 x \left(2 x + 7\right) = 0$

$6 x = 0$
$x = 0$

$2 x + 7 = 0$
$2 x = - 7$
$x = \frac{- 7}{2}$

Find the second derivative to judge for which value of "x" the function has a minimum, maximum and point of inflection.

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 24 x + 42$

At x=0; (d^2y)/(dx^2)=24(0)+42=42>0

At x=0 ; dy/dx=0; (d^2y)/(dx^2)>0
Hence the function has a minimum at $x = 0$

At x=(-7)/2; (d^2y)/(dx^2)=24((-7)/2)+42= -84+42=-42>0

At x=(-7)/2 ; dy/dx=0; (d^2y)/(dx^2)<0
Hence the function has a maximum at $x = \frac{- 7}{2}$

To find the point of inflection, set the second derivative equal to zero

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = 0 \implies 24 x + 42 = 0$
$x = \frac{- 42}{24}$
At $x = \frac{- 42}{24}$ there is a point of inflection