# For what values of x is f(x)= -x^3+3x^2-2x+2  concave or convex?

Nov 19, 2017

Concave (Convex) Up on the interval $\left(- \infty , 1\right)$
Concave (Convex) Down on the interval $\left(1 , \infty\right)$

#### Explanation:

We are given the function $f \left(x\right) = - {x}^{3} + 3 {x}^{2} - 2 x + 2$

$\textcolor{red}{S t e p .1}$

Find the First Derivative

$f ' \left(x\right) = - 3 {x}^{2} + 6 x - 2$

$\textcolor{red}{S t e p .2}$

Find the Second Derivative

$f ' ' \left(x\right) = - 6 x + 6$

$\textcolor{red}{S t e p .3}$

Next, set

$f ' ' \left(x\right) = - 6 x + 6 = 0$

Simplifying, we get $x = 1$

$\textcolor{red}{S t e p .4}$

Then, we consider a number larger than 1 and a number smaller than 1 and substitute the values in our Second Derivative.

If the number is Greater than 1, our f''(x) = -6x + 6" will yield a "Negative" number.

If the number is Less than than 1, our f''(x) = -6x + 6" will yield a "Positive" number.

Hence, we observe that $f \left(x\right)$ is "Concave Up" on the interval $\left(- \infty , 1\right)$ and "Concave Down" on the interval $\left(1 , \infty\right)$

Refer to the Number Line as shown below:

$$                                                    1


:..........................................................*..................................................................:

$$             Positive                                                  Negative