For what values of x is f(x)= 7x^3 + 2 x^2 + 7x -2 f(x)=7x3+2x2+7x2 concave or convex?

1 Answer
Mar 26, 2017

Concave down for x < -2/21x<221, concave up for x > -2/21x>221

Explanation:

First, we can try to find inflection points for this function. An inflection point is a point where the concavity changes, so finding this point is often helpful when analyzing concavity.

The process is straightforward; we will find xx such that d^2/dx^2 f(x) = 0d2dx2f(x)=0, that is, when the second derivative of ff is zero.

Start by finding the 1st derivative, by simply applying the power rule to each term:

d/dx f(x) = 21x^2 + 4x + 7ddxf(x)=21x2+4x+7

Then, differentiate again to find the 2nd derivative:

d^2/dx^2 f(x) = 42x + 4d2dx2f(x)=42x+4

So, now we set the thing equal to zero:

0 = 42x + 40=42x+4

And solve for xx:

x = -2/21x=221

So now we know that the function has one inflection point. What about the rest of possible values for xx?

Well, if a segment of a graph is concave up (its slope is increasing) then the 2nd derivative will be positive. And if a segment is concave down, with a decreasing slope, the 2nd derivative will be negative.

Since we know that the 2nd derivative switches from negative to positive or vice versa at x = -2/21x=221, let's see whether it's positive or negative, at, for instance, x = 0x=0:

d^2/dx^2 f(0) = 42*0 + 4 = 4d2dx2f(0)=420+4=4

Interesting. So the 2nd derivative is positive at x=0x=0. This tells us that, since the only switch occurs at x = -2/21x=221, all x > -2/21x>221 have a positive 2nd derivative as well, and are therefore concave up.

On the other hand, all x < -2/21x<221 are concave down, since the switch must occur at x=-2/21x=221.

Hopefully this makes sense.