Determine the intervals on which the function 𝑓𝑥 equals 𝑥 cubed minus 11 𝑥 plus two is concave up and down.
Okay, so before we can actually solve this problem, we need to actually understand what concave up and concave down mean. Well, in my sketch, I’ve actually drawn part of the function. What highlighted is that actually in this function the slope is increasing on this particular part of the function. That’s because we’re actually concave up.
However, in the second sketch, we can actually see if it’s concave down, so if the part of the function we’re looking at is concave down, then the slope is actually decreasing. So it’s actually slope which is the key term here, because actually when we’re looking at slope, we want to find out whether increases or decreases. And to enable us to do that, it tells us what we need to do next with this problem, because we need to find the slope function or 𝑑𝑦 𝑑𝑥.
So in order to find this, we’re actually gonna find the differential of 𝑓𝑥 and I’ve written that with with 𝑓 dash 𝑥. Actually, you can see that that is the same as 𝑑𝑦 𝑑𝑥. But this is just another way that actually we can notate that. So now to differentiate our function, we’re gonna get three 𝑥 squared minus 11.
And we get that because if we multiply our coefficient of 𝑥 cubed, which is just one, by our exponent, which is three, we get three. And then we reduce the exponent by one. It gives us two because it goes from three to two. So we get three 𝑥 squared. And then negative 11𝑥 just differentiates to negative 11.
Brilliant! So now we’ve actually got a slope function. But why is this useful? So this could be useful if we’re actually wanting to find the slope of our graph at certain points. However, for this question, we actually want to know where’s our graph concave up or concave down, so where is it increasing or decreasing with regards to its slope.
As we actually want to find out where the function is concave up and concave down, I’ve drawn three more sketches to kind of help us understand what to do next. The first sketch is actually a function. So I’ve just chosen a cubic function because our function in the question is cubic.
As I said, the first graph actually shows the cubic function. Second graph shows the actual differential of it. So it actually shows the slope function. And what we can see is that, in the bit that I’ve highlighted in pink, the slope is actually decreasing up until the point that I’ve marked with a spot. And this is further confirmed by our third graph, which is actually of the second differential, because we can see that, well, the second differential is negative up until this point. So thus, actually, the slope is decreasing.
We can actually see that, at the second part of this function, our slope is actually increasing. And now we can see that if we look at the first differential, so our slope function, which is the second graph, and our third graph, which is our second differential, this whole section is positive.
So as we’re trying to find out the interval where the slope is decreasing or slope is increasing, so concave up and concave down, there’s a key point. And that key point I’ve circled in orange. If you look at the first, second, and third graphs, it’s the third graph. We can see it’s the point where our second differential is equal to zero because this is where the slope goes from decreasing to increasing.
So in order to find that value, let’s set our second differential equal to zero. But in order to actually set our second differential to zero, we’re first gonna have to find it. And we do that by differentiating our first differential, which is three 𝑥 squared minus 11. When we differentiate that, we’re just left with six 𝑥.
So now what we’re gonna do is find the 𝑥 value when our second differential is equal to zero, the point that we highlighted on our graph. So we have zero is equal to six 𝑥. Okay, now to solve this, well, we can see clearly 𝑥 is equal zero. And this is the key point because it’s the point that we marked on the graph as I said. And it’s actually going to be our point of inflection.
So this is where we’re gonna change between our concave up and concave down. So now actually if we consider what we’ve got, we’ve got our zero point. So that’s our point of inflection. And then we can say that everything to the left up until negative infinity is going to be concave down. So that’s actually where our slope is decreasing. And everything to the right is going to be positive, so concave up, because that’s where our slope is increasing.
However, we want to check this. And to do that, we’re actually gonna substitute in values either side of our zero to see whether they actually fit with what we think is concave up and concave down.
So I’m gonna start by substituting 𝑥 equals one into our second derivative of our function. So we’re gonna get that the second derivative is equal to six multiplied by one, which gives us the answer six, which is positive. So great! That actually shows that that’s correct. So our inflection point zero seems to work there because a number just to the right of it is positive. So that’s going to be concave up. And the slope is increasing.
So now we’re actually gonna substitute in 𝑥 is equal to negative one because this is a number less than our zero value. So therefore our second differential is equal to six multiplied by negative one, which gives us negative six, which is also a negative value.
So concave down is actually gonna be in the interval negative infinity to zero. So you see that everything less than zero is actually gonna give us a point where the slope is decreasing. So therefore we’re gonna get concave down and concave up is gonna be in the interval zero to infinity cause we can actually see that everything on the right of our zero point, so our point of inflection, is actually gonna be where the slope is increasing, so therefore concave up.