Question

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

Answer 1

`f(x)` is convex (concave up) for `x in ("–"oo," "4/27)`.
`f(x)` is concave (concave down) for `x in (4/27," "oo)`.

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(x)="–"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(x)=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''(x)`.

Given `f(x)="–"9x^3+4x^2+7x-2`, we have

`f'(x)="–"27x^2+8x+7`

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

`f''(x)="–"54x+8`.

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

`f''(x)="–"54x+8 > 0`

`=>"                 –"54x>"–8"`

`=>"                       "x<8/54=4/27`.

So `f(x)` is convex (concave up) for `x in ("–"oo," "4/27)`. This means `x=4/27` is an inflection point—a point at which the direction of concavity changes.

Finding where `f(x)` is concave (concave down) means finding where `f''(x)<0`, but that's equivalent to swapping `>` signs for`<` signs from when we found `f''(x) > 0`. So our answer will be `f(x)` is concave for `x in (4/27," "oo)`.

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 `"–"oo`, 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.

Answer 2

`f(x)` 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(x)=-9x^3+4x^2+7x-2`

`f'(x)=-27x^2+8x+7`

To determine the critical points, we solve the equation

`-27x^2+8x+7=0`

`Delta=8^2-4*(-27)*(7)=820`

As, `Delta>0`, there are 2 real roots

`x_1=(-8+sqrt820)/(2*-27)=-0.382`

`x_2=(-8-sqrt820)/(2*-27)=0.678`

The chart of variations is

`color(white)(aaaa)` `x` `color(white)(aaaaaa)` `-oo` `color(white)(aaa)` `-0.382` `color(white)(aaaa)` `0.678` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x-x_1` `color(white)(aaaaaa)` `+` `color(white)(aaaaaaa)` `-` `color(white)(aaaaaa)` `-`

`color(white)(aaaa)` `x-x_2` `color(white)(aaaaaa)` `-` `color(white)(aaaaaaa)` `-` `color(white)(aaaaaa)` `+`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaaaa)` `-` `color(white)(aaaaaaa)` `+` `color(white)(aaaaaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `↘` `color(white)(aaaaaaa)` `↗` `color(white)(aaaaaa)` `↘`

So,

`f(x)` is convex when `x in ]-oo,0.678]`

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