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# For what values of x is f(x)= -9x^3 + 4 x^2 + 7x -2  concave or convex?

Feb 18, 2017

$f \left(x\right)$ is convex (concave up) for $x \in \left(\text{–"oo," } \frac{4}{27}\right)$.
$f \left(x\right)$ is concave (concave down) for $x \in \left(\frac{4}{27} , \text{ } \infty\right)$.

#### Explanation:

A function is concave (or concave down) where its derivative is decreasing. Graphically, a concave region looks like a cave (or cut from a cave shape). (Quick example: the function $f \left(x\right) = \text{–} {x}^{2}$ is concave everywhere, since it never curves upward.)

A function is convex (or concave up) where its derivative is increasing. Graphically, a convex region looks like a V (or cut from a V shape). (The function $f \left(x\right) = {x}^{2}$ is convex everywhere.)

To find these regions, we need to analyze the behaviour of the function's derivative. Hence, we need to find the derivative of the derivative: $f ' ' \left(x\right)$.

Given $f \left(x\right) = \text{–} 9 {x}^{3} + 4 {x}^{2} + 7 x - 2$, we have

$f ' \left(x\right) = \text{–} 27 {x}^{2} + 8 x + 7$

by the power rule, and so the derivative of this is

$f ' ' \left(x\right) = \text{–} 54 x + 8$.

Just like how we know $f \left(x\right)$ is increasing when $f ' \left(x\right) > 0$, we know $f ' \left(x\right)$ is increasing when $f ' ' \left(x\right) > 0$. Using this, we find:

$f ' ' \left(x\right) = \text{–} 54 x + 8 > 0$

$\implies \text{ –"54x>"–8}$

$\implies \text{ } x < \frac{8}{54} = \frac{4}{27}$.

So $f \left(x\right)$ is convex (concave up) for $x \in \left(\text{–"oo," } \frac{4}{27}\right)$. This means $x = \frac{4}{27}$ is an inflection point—a point at which the direction of concavity changes.

Finding where $f \left(x\right)$ is concave (concave down) means finding where $f ' ' \left(x\right) < 0$, but that's equivalent to swapping $>$ signs for$<$ signs from when we found $f ' ' \left(x\right) > 0$. So our answer will be $f \left(x\right)$ is concave for $x \in \left(\frac{4}{27} , \text{ } \infty\right)$.

Here's a graph of the function, with the inflection point circled.
graph{(-9x^3+4x^2+7x-2-y)((x-4/27)^2+(y+0.9)^2/36-0.0025)=0 [-2.1, 2.1, -6, 6]}

As we come from $\text{–} \infty$, the slope (of the tangent line) of the graph is increasing, right up until the inflection point, where the slope begins to decrease.

Notice how the function is V-shaped to the left of the inflection point, and cave-shaped to its right. That's an easy way to remember the difference between concave and convex: concave is like a cave; convex is like a V.

Feb 18, 2017

$f \left(x\right)$ is convex when x in ]-oo,0.678] and concave when x in [-0.382,+oo[

#### Explanation:

We calculate the first derivative and build a chart of variations

$f \left(x\right) = - 9 {x}^{3} + 4 {x}^{2} + 7 x - 2$

$f ' \left(x\right) = - 27 {x}^{2} + 8 x + 7$

To determine the critical points, we solve the equation

$- 27 {x}^{2} + 8 x + 7 = 0$

$\Delta = {8}^{2} - 4 \cdot \left(- 27\right) \cdot \left(7\right) = 820$

As, $\Delta > 0$, there are 2 real roots

${x}_{1} = \frac{- 8 + \sqrt{820}}{2 \cdot - 27} = - 0.382$

${x}_{2} = \frac{- 8 - \sqrt{820}}{2 \cdot - 27} = 0.678$

The chart of variations is

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a}$$- 0.382$$\textcolor{w h i t e}{a a a a}$$0.678$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x - {x}_{1}$$\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}$$-$$\textcolor{w h i t e}{a a a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$x - {x}_{2}$$\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}$$-$$\textcolor{w h i t e}{a a a a a a}$$+$

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

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a}$↘$\textcolor{w h i t e}{a a a a a a a}$↗$\textcolor{w h i t e}{a a a a a a}$↘

So,

$f \left(x\right)$ is convex when x in ]-oo,0.678]

and concave when x in [-0.382,+oo[