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

1 Answer
Mar 26, 2017

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

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 #x# such that #d^2/dx^2 f(x) = 0#, that is, when the second derivative of #f# is zero.

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

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

Then, differentiate again to find the 2nd derivative:

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

So, now we set the thing equal to zero:

#0 = 42x + 4#

And solve for #x#:

#x = -2/21#

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

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 = -2/21#, let's see whether it's positive or negative, at, for instance, #x = 0#:

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

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

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

Hopefully this makes sense.