Question Video: Finding Intervals of Upward and Downward Concavity of a Polynomial Mathematics • Higher Education

Determine the intervals on which the function 𝑓(π‘₯) = βˆ’4π‘₯⁡ + π‘₯Β³ is convex up and down.

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Video Transcript

Determine the intervals on which the function 𝑓 of π‘₯ equals negative four π‘₯ to the fifth power plus π‘₯ cubed is convex up and down.

Our function 𝑓 of π‘₯ is a fifth-degree polynomial, and we’re looking to determine its convexity. We know that we can use the second derivative to do so. And in fact, since 𝑓 of π‘₯ is a polynomial, this means it’s differentiable and continuous over its entire domain. This means we can be confident that we can evaluate both the first and second derivative of this function and ensure that it exists.

So how does the second derivative help us identify the convexity of a function? Well, we know that the function is convex up over any intervals where the second derivative is less than zero, where it’s negative. And it’s convex down over intervals where the second derivative is positive, in other words, where the first derivative, the slope, is increasing. So let’s begin by evaluating the first then the second derivative of our function.

To begin, we’ll find the first derivative. And we know we can simply differentiate term by term. The power rule for differentiation says that we can multiply by the exponent and then reduce that exponent by one. So when we differentiate negative four π‘₯ to the fifth power with respect to π‘₯, we get five times negative for π‘₯ to the fourth power. Similarly, the derivative of π‘₯ cubed is three π‘₯ squared. And so our first derivative is negative 20π‘₯ to the fourth power plus three π‘₯ squared.

We’ll repeat this process for 𝑓 double prime of π‘₯. This time, differentiating term by term and simplifying gives us negative 80π‘₯ cubed plus six π‘₯. Now this expression for the second derivative is a cubic polynomial. It has a negative leading coefficient, which tells us information about the shape of the curve. This means, if we think about the curve in particular, we’ll be able to identify the regions where the second derivative is negative and where it’s positive.

We’ll begin by finding where it’s actually equal to zero. This will tell us the points where it intersects the π‘₯-axis, in other words, for what values of π‘₯ is negative 80π‘₯ cubed plus six π‘₯ equal to zero. Factoring the left-hand side, and we get two π‘₯ times negative 40π‘₯ squared plus three. And for the product of these two expressions to be zero, we know one or other of the expressions must itself be equal to zero. In other words, either two π‘₯ is equal to zero or negative 40π‘₯ squared plus three equals zero.

Dividing our first equation by two, and we get π‘₯ equals zero. With our second equation, we’ll add 40π‘₯ squared then divide through by 40. So π‘₯ squared is three over 40. To solve for π‘₯, we take the positive and negative square root of three over 40. And π‘₯ simplifies to positive or negative root 30 over 20. So this tells us that our cubic curve negative 80π‘₯ cubed plus six π‘₯ passes through the π‘₯-axis at zero and positive and negative root 30 over 20. Since it has a negative leading coefficient, we also know that it should look like this.

So with that in mind, let’s identify the regions over which this second derivative is negative. This will tell us where the original function 𝑓 of π‘₯ is convex up. Specifically, we see that the second derivative is less than zero. In other words, the graph of this function lies below the π‘₯-axis over the open interval negative root 30 over 20 to zero and the open interval root 30 over 20 to ∞. So these are the regions over which it’s convex up.

With that in mind, let’s repeat this process and identify the portions where 𝑓 double prime of π‘₯ is positive. On the graph of 𝑦 equals 𝑓 double prime of π‘₯, these are the regions where it lies above the π‘₯-axis. So we see that 𝑓 double prime of π‘₯ is greater than zero over the open interval from negative ∞ to negative root 30 over 20 and the open interval from zero to root 30 over 20. This means these are the regions over which the original function is convex down.

And so we’ve answered the question. The function 𝑓 of π‘₯ is convex up over the open interval negative root 30 over 20 to zero and the open interval root 30 over 20 to ∞. And it’s convex down over the open interval negative ∞ to negative root 30 over 20 and zero to root 30 over 20.

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