# Find the open intervals where the function is concave up and concave down. y=-x^3+2x^2-2

## Question:

Find the open intervals where the function is concave up and concave down.

{eq}y=-x^3+2x^2-2 {/eq}

## Concavity of a Function

{eq}{/eq}

Consider a function given by :

$$\displaystyle y = f(x) \\$$

1. f(x) is concave up if :

$$\displaystyle f''(x) = \frac{d^2y}{dx^2} \gt 0 \\$$

2. f(x) is concave down if :

$$\displaystyle f''(x) = \frac{d^2y}{dx^2} \lt 0 \\$$

{eq}{/eq}

Given function :

$$\displaystyle y=-x^3+2x^2-2 \\$$

To find the intervals in which the function is concave up or down, we need to obtain the second derivative of y with respect to x.

{eq}\displaystyle{\frac{dy}{dx} = \frac{d}{dx} (-x^3+2x^2-2 ) \\ \Rightarrow \frac{dy}{dx} = -\frac{d}{dx} (x^3) + 2\frac{d}{dx}(x^2) - \frac{d}{dx}(2) \\ \Rightarrow \frac{dy}{dx} = -3x^2 + 2(2x) + 0 \\ \Rightarrow \frac{dy}{dx} = -3x^2 + 4x \\ } {/eq}

{eq}\displaystyle{\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) \\ \Rightarrow \frac{d^2y}{dx^2} = \frac{d}{dx}(-3x^2 + 4x) \\ \Rightarrow \frac{d^2y}{dx^2} = -3 \frac{d}{dx}(x^2) + 4 \frac{d}{dx}(1) \\ \Rightarrow \frac{d^2y}{dx^2} = -3(2x) + 4 \\ \Rightarrow \frac{d^2y}{dx^2} = -6x + 4 \\ } {/eq}

For y to be concave up,

{eq}\displaystyle{\frac{d^2 y}{dx^2} \gt 0 \\ \Rightarrow -6x + 4 \gt 0 \\ \Rightarrow x \lt \frac{2}{3} \\ } {/eq}

$$\displaystyle \therefore \text{y is concave up for } x \ \in \ (-\infty \ , \ \frac{2}{3}) \\$$

For y to be concave down ,

{eq}\displaystyle{\frac{d^2 y}{dx^2} \lt 0 \\ \Rightarrow -6x + 4 \lt 0 \\ \Rightarrow x \gt \frac{2}{3} \\ } {/eq}

$$\displaystyle \therefore \text{y is concave down for } x \ \in \ (\frac{2}{3} \ , \ \infty) \\$$