# How do you determine whether the function f(x)= 5+12x- x^3 is concave up or concave down and its intervals?

Aug 27, 2015

Refer the explanation section

#### Explanation:

Given -
y = $- {x}^{3} + 12 x + 5$

Find the first 2 derivatives -

$\frac{\mathrm{dy}}{\mathrm{dx}}$ = $- 3 {x}^{2} + 12$
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$ = $- 6 x$

Set the 1st derivative to zero to find for what value of 'x' the curve turns.

$\frac{\mathrm{dy}}{\mathrm{dx}} \implies$ $- 3 {x}^{2} + 12$=0
$- 3 {x}^{2}$ = - 12
${x}^{2}$ = $\frac{- 12}{-} 3$= 4
x = $\pm \sqrt{4}$

x = 2
x = - 2

At x = 2 and x = - 2 the curve turns. To determine whether it turns upwards or downwards, substitute the values in the 2nd derivative.

At x = 2 ; $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$ = $- 6 x$ =$- 6 \times 2$= - 12 < 0

The curve has a maximum at x = 2. In the immediate proximity the curve is concave downwards.

At x = - 2 ; $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$ = $- 6 x$ =$- 6 \times - 2$= 12 > 0

The curve has a Minimum at x = - 2. In the immediate proximity the curve is concave upwards.

Point of inflection separates concavity from convexity. To the Point of inflection, set the 2nd derivative equal to zero.

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 0 \implies - 6 x = 0$
x = 0

At x = 0 , there is point of inflection.

-$\infty$ < x < 0 ; curve is concave upwards
$\infty$ > x > 0 ; curve is concave downwards.

graph{-x^3 + 12 x + 5 [-74.04, 74.1, -37.03, 37]}