Video: Finding the Unknown Coefficients in the Expression of a Function given the Value of the Second and Third Derivatives of a Function

Video Transcript

Given that 𝑦 is equal to 𝑎𝑥 cubed plus 𝑏𝑥 squared, the third derivative of 𝑦 is negative 18, and the second derivative of 𝑦 with respect to 𝑥 evaluated that 𝑥 equals two is negative 14, find 𝑎 and 𝑏.

Here, we have an equation for 𝑦 in terms of 𝑥 and some information about the second derivative and the third derivative, denoted as 𝑦 prime prime prime. To answer this question, let’s begin just by finding an equation for the second and third derivatives of 𝑦 with respect to 𝑥.

Differentiating 𝑦 with respect to 𝑥, and we get three 𝑎𝑥 squared plus two 𝑏𝑥. To find the second derivative, we’ll differentiate the equation for the second derivative. That’s two times three 𝑎𝑥 plus two 𝑏. That simplifies to six 𝑎𝑥 plus two 𝑏. Once again, to find the third derivative, we differentiate the second derivative with respect to 𝑥. Since two 𝑏 is a constant, the third derivative is six 𝑎.

We’re told that the second derivative evaluated at 𝑥 is equal to two is negative 14. So, let’s substitute 𝑥 is equal to two into our equation for the second derivative and set it equal to negative 14. That’s six times two plus two 𝑏 equals negative 14, or 12𝑎 plus two 𝑏 is negative 14. Similarly, we’re told that the third derivative is equal to negative 18. So, we can say that six 𝑎 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 𝑎 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 𝑏 equals negative 14. 12 multiplied by negative three is negative 36. We add 36 to both sides of our equation to get two 𝑏 equals 22. And we divide through by two to get 𝑏 equals 11. 𝑎 equals negative three and 𝑏 equals 11.