### Video Transcript

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

Our function π of π₯ is a
fifth-degree polynomial, and weβre looking to determine its convexity. We know that we can use the second
derivative to do so. And in fact, since π of π₯ is a
polynomial, this means itβs 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
itβs negative. And itβs convex down over intervals
where the second derivative is positive, in other words, where the first derivative,
the slope, is increasing. So letβs begin by evaluating the
first then the second derivative of our function.

To begin, weβll 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 four 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.

Weβll repeat this process for π
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, weβll be able to identify the regions where the second
derivative is negative and where itβs positive.

Weβll begin by finding where itβs
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, weβll 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, letβs
identify the regions over which this second derivative is negative. This will tell us where the
original function π 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 β. So these are the regions over which
itβs convex up.

With that in mind, letβs repeat
this process and identify the portions where π double prime of π₯ is positive. On the graph of π¦ equals π double
prime of π₯, these are the regions where it lies above the π₯-axis. So we see that π double prime of
π₯ is greater than zero over the open interval from negative β 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 weβve answered the
question. The function π 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 β. And itβs convex down over the open
interval negative β to negative root 30 over 20 and zero to root 30 over 20.