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# How do you determine the intervals where f(x)=3x^2-x^3/3 is concave up or down?

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#### Explanation

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Ben Share
Jun 19, 2018

Determine where the second derivative is positive or negative. If negative, then it is concave down. If positive, concave up. If zero, then it's an inflection point.

#### Explanation:

While the first derivative tells us the slope of a function at a given point, the second derivative gives us the concavity of the function.

When the second derivative is positive, the function has a 'concave up' concavity to it. When negative, it's concave down. The point where this changes is the point of inflection. The point of inflection is equal to when the second derivative is equal to zero.

Let's work with the function for a bit to determine the second derivative:

$f \left(x\right) = 3 {x}^{2} - \frac{{x}^{3}}{3}$

$f ' \left(x\right) = \textcolor{red}{2} \cdot 3 x - \textcolor{red}{3} \frac{{x}^{\textcolor{red}{2}}}{3}$

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

$f ' ' \left(x\right) = 6 - \textcolor{red}{2} x$

Now that we know the second derivative, we can calculate the points of inflection to determine the intervals for concavity:

$f ' ' \left(x\right) = 0 = 6 - 2 x$

$2 x = 6$

$x = 3$

We only have one inflection point, so we just need to determine if the function is concave up or down on either side of the function:

$f ' ' \left(2\right) = 6 - 2 \left(2\right)$

$f ' ' \left(2\right) = 2 \therefore \text{concave up}$

$f ' ' \left(4\right) = 6 - 2 \left(4\right)$

$f ' ' \left(2\right) = - 2 \therefore \text{concave down}$

The concavity for our function can be described as such:

$x < 3$ Concave up
$x > 3$ Concave down

We can see that when graphing the function as well:

graph{3x^2-(x^3)/3 [-5, 10, -2, 38]}

Then teach the underlying concepts
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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Jun 19, 2018

The function is concave up in the interval $\left(- \infty , 3\right)$ and concave down in the interval $\left(3 , + \infty\right)$

#### Explanation:

The function is

$f \left(x\right) = 3 {x}^{2} - {x}^{3} / 3$

This a polynomial function continous and derivable on $\mathbb{R}$.

Calculate the first and second derivatives

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

$f ' ' \left(x\right) = 6 - 2 x$

Let $f ' \left(x\right) = 0$

$\implies$, $6 - 2 x = 0$

$\implies$, $x = 3$

There is an inflection point at $\left(3 , 18\right)$

Build a variation chart to determine the concavities

$\textcolor{w h i t e}{a a a a}$$\text{ Interval }$$\textcolor{w h i t e}{a a a a}$$\left(- \infty , 3\right)$$\textcolor{w h i t e}{a a a a}$$\left(3 , + \infty\right)$

$\textcolor{w h i t e}{a a a a}$$\text{ sign f''(x) }$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$\text{ f(x) }$$\textcolor{w h i t e}{a a a a a a a a a a a}$$\cup$$\textcolor{w h i t e}{a a a a a a a a a}$$\cap$

The function is concave up in the interval $\left(- \infty , 3\right)$ and concave down in the interval $\left(3 , + \infty\right)$

graph{3x^2-x^3/3 [-9.73, 18.75, -0.11, 14.13]}

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