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Current time:0:00Total duration:6:11

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.4 (EK)

, FUN‑4.A.5 (EK)

, FUN‑4.A.6 (EK)

- [Instructor] Robert was
asked to find where g of x, which is equal to the cube root
of x, has inflection points. This is his solution. And then later we are asked,
"Is Robert's work correct? "If not, what's his mistake?" So pause this video and try
to figure it out on your own. All right, now let's work
through this together. So our original g of x is
equal to the cube root of x, which is the same thing as x to the 1/3. So in step one it looks
like Robert's trying to find the first and second derivative. So the first derivative,
we just do the power rule, so it'll be 1/3x to the
decrement of the exponent, so this is looking good. Second derivative, we take
this, multiply this times 1/3, which would be negative 2/9. And then decrement, negative 2/3, which would indeed by negative
5/3, so that looks right. And then it looks like
Robert's trying to rewrite it. So we have the negative 2/9 still, but then he recognized
that this is the same thing as x to 5/3 in the denominator, and x to the 5/3 is the same thing as the cube root of x to the fifth. So this is all looking good. Step one looks good. And then step two, it
looks like he's trying to find the solution or
he's trying to find x values where the second derivative
is equal to zero, and it is indeed true
that this has no solution, that you can never make
this second derivative equal to zero. In order to be zero, the
numerator would have to be zero. And, well, two is never
going to be equal to zero. So this is correct. And then step three, he says g doesn't have
any inflection points. Now, this is a little bit suspect. It is in many cases our
inflection point is a situation where our second derivative
is equal to zero, and even then we don't know
it's an inflection point. It would be a candidate inflection point. We would have to confirm that our second derivative crosses signs or switches signs as
we cross that x value. But here we can't find a situation where our second derivative
is equal to zero, but we have to remind ourselves that other candidate inflection points are where our second
derivative is undefined. And so he can't make this
statement without seeing where our second derivative
could be undefined. So, for example, he could say that g prime prime is undefined when what? Well, this is going to be undefined when x is equal to zero. Zero to the fifth, cube root
of that, that's gonna be zero. But then you're dividing by zero. So g prime prime undefined
when x is equal to zero. So therefore x equals,
so we could say candidate inflection point when x equals zero. And so then we would want to test it. And we could set up a traditional table that you might have seen before where we have our interval or intervals. We could have test
values in our intervals. We have to be careful with those. Make sure that they are indicative. And then we would say the sign of our second derivative of g prime prime. And then we would have our concavity. Concavity of g. And in order for x equals zero
to be an inflection point, we would have to switch signs, or our second derivative
would have to switch signs as we cross x equals zero, which would mean our
concavity of g switches signs as we cross x equals zero. So let's do values less than zero, negative infinity to zero and then values greater
than zero, zero to infinity. I could do test values. Let's say, I'll use negative one and one. And you have to be careful
when you use these. You have to make sure
that we are close enough that nothing unusual happens
between these test values up until we get to that
candidate inflection point. And, now, what's the sign
of our second derivative when x is equal to negative one? When x equals negative one. So, let's see, negative one to the fifth power is negative one. Cube root of negative one is negative one. And so we're gonna have negative 2/9 divided by negative one. It's gonna be positive 2/9. So our sign right over
here is gonna be positive. And this is gonna be in general when we're dealing with
any negative value, 'cause if you take any negative
value to the fifth power, it's gonna be negative. And you take the cube root of that and you're gonna have negative. But then you have a negative
value divided by that. You're gonna get a positive value. So you can feel good that
this test value's indicative of actually this entire interval. And if you're dealing
with a positive value, well, that to the fifth
power is gonna be positive. Cube root of that is still
going to be positive. But then you're gonna have negative 2/9 divided by that positive value, so this is going to be negative. So it is indeed the case that
our concavity of g switches as we cross x equals zero. We're concave upwards
when x is less than zero. Our second derivative is positive. And we're concave downwards
when x is greater than zero. Let me write that a little bit. Downwards. Downwards when x is greater than zero. So we are switching concavity
as we cross x equals zero, and so this tells us that x. So let's see, we are switching signs. Switching. Let me say g prime prime switching signs as we cross x equals zero. And our function is
defined at x equals zero. And function defined at x equals zero. So we have an inflection
point at x equals zero. So inflection point at x is equal to zero. And if you're familiar with
the graph of the cube root, you would indeed see an
inflection point at that point. So there we go. He was wrong in step three. There actually is an inflection point. It's not when the second
derivative is equal to zero. It's actually where the second
derivative is undefined.

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