### Video Transcript

Find, if any, the inflection points
of π of π₯ equals three π₯ squared times the natural log of two π₯.

To find the points of inflection,
weβll evaluate the second derivative of our function and set it equal to zero. Notice that our function is itself
the product of two functions. So weβll need to use the product
rule to differentiate it. This says that for two
differentiable functions, π’ and π£, the derivative of their product is π’ times dπ£
by dπ₯ plus π£ times dπ’ by dπ₯. We let π’ be equal to three π₯
squared and π£ be equal to the natural log of two π₯. Then dπ’ by dπ₯ is six π₯ and dπ£
by dπ₯ is one over π₯. So π prime the first derivative of
our function is three π₯ squared times one over π₯ plus six π₯ times the natural log
of two π₯ or three π₯ plus six π₯ times the natural log of two π₯. Weβre going to differentiate this
again.

We use the product rule to find
that the derivative of six π₯ times the natural log of two π₯ is six plus six times
the natural log of two π₯. And the second derivative is nine
plus six times the natural log of two π₯. Letβs set this equal to zero. To solve for π₯, we subtract nine
and then divide through by six. We raised both sides as the power
of π. And then we divide through by
two. So potentially, thereβs an
inflection point at π₯ equals one-half π to the negative three over two. But we ought to check whether this
is actually an inflection point by checking the values of π double prime or the
second derivative to either side of this.

A half π to the negative three
over two is approximately 0.112. So letβs try π₯ equals 0.1 and π₯
equals 0.12. The second derivative π double
prime of 0.1 is less than zero. And π double prime of 0.12 is
greater than zero. The curve goes from being concave
down to being concave up. And we can say we do indeed have a
point of inflection at π₯ equals one-half π to the negative three over two. Substituting this value of π₯ into
original function, we get negative nine over eight π cubed. π has an inflection point at π to
the negative three over two over two, negative nine over eight π cubed.