Hi everyone!
Welcome back to integralcalc.com.
Today we’re going to be doing a problem with the second derivative test.
In this particular problem, we’ve been asked to use the second derivative test to determine
the nature of the critical points f(x) = x^2 – 3x + 1.
Remember that the second derivative test, which I’ve pretty much summarized right
here, tells us that if we find the critical points of a function and then we evaluate
them in a second derivative, if we get a positive result or a result that’s greater than zero,
we know that that critical point represents at least a local minimum of the function.
And if we get an answer that’s negative or less than zero, we know that the critical
point represents a local maximum.
What we’re going to do in order to use the second derivative test to evaluate the critical
points are the following three steps.
We’re going to find the first derivative f’(x) and then the critical points.
Then we’ll find the second derivative f’’(x) and then we will plug our critical points
into the second derivative to define the nature of each of our critical points.
Proceeding here with step one, the first thing we’re going to do is find the first derivative.
f’(x), this function is pretty basic.
We’re just using this function to illustrate the second derivative test.
So we’ll use the power rule to evaluate each of these terms.
We’ll get 3x^2 – 3.
The derivative of 1 of course is zero because it’s a constant.
So this is our first derivative, 3x^2 – 3.
Now in order to find critical points, remember that we have to set the frist derivative equal
to zero and solve for x.
So we’ll say 0 = 3x^2 – 3 and now we want to solve for x. the way that we’ll do that
is by factoring the right hand side.
So we’re going to factor out a 3 and we’ll get 3(x^2 - 1).
Remember also that we want to factor as completely as possible and we can still factor x^ - 1
so we’ll go ahead and do that.
We’ll get 3(x + 1)(x - 1).
Now that we’ve factored as completely as possible, we can define the critical points
of f(x) so we’ll solve each of these pieces for x.
We know that if we set this piece equal to zero, we’ll get x = -1 ad if we set this
piece equal to zero, we’ll get x = 1.
We have two critical points of the function, -1 and 1.
That’s the end of step 1.
Step two is for us to find the second derivative.
We already know that our first derivative is 3x^2 – 3.
We’ll use power rule again to find our second derivative.
We’ll call it f”(x).
We’re going to end up with 6x.
The 3 here goes away because it’s a constant.
The derivative of any constant is 0.
So our second derivative or f”(x) = 6x.
Now in order to define the nature of each of our critical points, we’re going to plug
in our critical points to our second derivative function.
So we’re plugging both of these into our second derivative.
Let’s go ahead first and plug in -1 so we’ll find f”(-1).
That’ll be 6(-1) = -6.
-6 is clearly less than zero, which means if we go back and look at our definition,
that this here is true which means that the critical point x = -1 is going to represent
a local maximum.
It may represent a global max but we would have to determine where the function is increasing
and decreasing and the actual value of each critical point in order to determine whether
or not it was truly a global maximum.
So now, we’ll find f” of the second critical point.
6(1) = 6 which is obviously greater than zero and according to the definition of the second
derivative here, when it’s greater than zero, we’re looking at a local minimum.
Again, a possibly a global minimum but we’d have to evaluate in order to determine that
for sure.
So we have a maximum and a minimum.
It’s kind of difficult to remember because it’s opposite.
Obviously if the result is less than zero or negative, it’s maximum.
If it’s greater than zero, it’s a minimum.
Just remember that it’s opposite in that way.
But that is how you use the second derivative test to determine whether or not your critical
points represent maximums or minimums.
So that’s it.
I hope this video helped you guys and I will see you in the next one.