### Video Transcript

Given that π¦ is equal to ππ₯
cubed plus ππ₯ squared, the third derivative of π¦ is negative 18, and the second
derivative of π¦ with respect to π₯ evaluated that π₯ equals two is negative 14,
find π and π.

Here, we have an equation for π¦ in
terms of π₯ and some information about the second derivative and the third
derivative, denoted as π¦ prime prime prime. To answer this question, letβs
begin just by finding an equation for the second and third derivatives of π¦ with
respect to π₯.

Differentiating π¦ with respect to
π₯, and we get three ππ₯ squared plus two ππ₯. To find the second derivative,
weβll differentiate the equation for the second derivative. Thatβs two times three ππ₯ plus
two π. That simplifies to six ππ₯ plus
two π. Once again, to find the third
derivative, we differentiate the second derivative with respect to π₯. Since two π is a constant, the
third derivative is six π.

Weβre told that the second
derivative evaluated at π₯ is equal to two is negative 14. So, letβs substitute π₯ is equal to
two into our equation for the second derivative and set it equal to negative 14. Thatβs six times two plus two π
equals negative 14, or 12π plus two π is negative 14. Similarly, weβre told that the
third derivative is equal to negative 18. So, we can say that six π must be
equal to negative 18.

Notice, that this latter equation
has a single variable, so we can solve as normal. We can divide both sides of this
equation by six. And when we do, we see that π is
equal to negative three. We can take this value and
substitute it into the equation we formed using the second derivative. That gives us 12 multiplied by
negative three plus two π equals negative 14. 12 multiplied by negative three is
negative 36. We add 36 to both sides of our
equation to get two π equals 22. And we divide through by two to get
π equals 11. π equals negative three and π
equals 11.