### Video Transcript

The function ℎ of 𝑥 which is equal to 𝑥 cubed plus 𝑘𝑥 squared multiplied by 𝑒 to the power of negative 𝑥 has a critical point at 𝑥 equals one. Find 𝑘 and list all the critical points.

The question gives us a function ℎ of 𝑥 which is a product of a polynomial function and an exponential function. And we are told that this function has a critical point when 𝑥 equals one. We are asked to find the constant 𝑘 together with all the critical points of the function. We begin by recalling that 𝑥 equals 𝑎 is a critical point of ℎ of 𝑥 if ℎ prime of 𝑎 equals zero or ℎ prime of 𝑎 does not exist, where ℎ prime of 𝑎 is the derivative of ℎ evaluated at 𝑥 equals 𝑎. Since we have a product of a polynomial function and an exponential function and both of these are differentiable for all real values, then the product will be differentiable for all real values. This means that ℎ prime of 𝑥 will exist for all real values. And we therefore just need to find the values of 𝑎 such that ℎ prime of 𝑎 is equal to zero.

Since we are told there is a critical point at 𝑥 equals one, we know that ℎ prime of one must equal zero. To find any further critical points of ℎ of 𝑥, we need to find an expression for ℎ prime of 𝑥. Since ℎ of 𝑥 is the product of two functions, we’ll differentiate this using the product rule of differentiation. This tells us that the derivative of the product of two functions 𝑢 and 𝑣 is equal to 𝑢 multiplied by 𝑣 prime plus 𝑣 multiplied by 𝑢 prime. We will set 𝑢 equal to 𝑥 cubed plus 𝑘𝑥 squared and 𝑣 equal to 𝑒 to the power of negative 𝑥. So ℎ of 𝑥 is equal to the product of 𝑢 of 𝑥 and 𝑣 of 𝑥.

We now need to find expressions for 𝑢 prime of 𝑥 and 𝑣 prime of 𝑥. We can differentiate the polynomial 𝑥 cubed plus 𝑘𝑥 squared term by term using the power rule of differentiation; we multiply it by the exponent of 𝑥 and then reduce the exponent by one. This gives us three 𝑥 squared plus two 𝑘𝑥. We can now find an expression for 𝑣 prime of 𝑥, that is, the derivative of 𝑒 to the power of negative 𝑥 with respect to 𝑥. We can differentiate this using our derivative rules for exponential functions. For any constant 𝑛, the derivative of 𝑒 to the power of 𝑛𝑥 is equal to 𝑛 multiplied by 𝑒 to the power of 𝑛𝑥. In our case, 𝑛 is equal to negative one. So 𝑣 prime of 𝑥 is equal to negative 𝑒 to the power of negative 𝑥.

We are now in a position to find ℎ prime of 𝑥 using the product rule, it is equal to 𝑢 of 𝑥 multiplied by 𝑣 prime of 𝑥 plus 𝑣 of 𝑥 multiplied by 𝑢 prime of 𝑥. Substituting in our expressions, we have the following: 𝑥 cubed plus 𝑘𝑥 squared multiplied by negative 𝑒 to the power of negative 𝑥 plus 𝑒 to the power of negative 𝑥 multiplied by three 𝑥 squared plus two 𝑘𝑥. We can simplify this expression by taking out a common factor of 𝑒 to the power of negative 𝑥 together with a common factor of 𝑥. This gives us ℎ prime of 𝑥 is equal to 𝑥𝑒 to the power of negative 𝑥 multiplied by negative 𝑥 squared minus 𝑘𝑥 plus three 𝑥 plus two 𝑘.

We remember that ℎ prime of 𝑥 must be equal to zero at the critical point. We therefore have the product of three factors equal zero. This means that one of these factors must be zero. Since our first factor is just 𝑥. We know that if 𝑥 is equal to zero, ℎ prime of 𝑥 equals zero. This means that 𝑥 equals zero is a critical point. We also know that 𝑒 is a positive number, so 𝑒 raised to any power will be positive. This tells us that 𝑒 to the power of negative 𝑥 is never equal to zero. And as a result, any remaining critical points must be roots of our quadratic factor.

In particular, we are told that the function has a critical point at 𝑥 equals one. This means that 𝑥 equals one is a root of the quadratic. Substituting 𝑥 equals one and setting the quadratic equal to zero, we have zero is equal to negative one squared minus 𝑘 multiplied by one plus three multiplied by one plus two 𝑘. The right-hand side simplifies to negative one minus 𝑘 plus three plus two 𝑘, which in turn gives us 𝑘 plus two. And this is equal to zero. Subtracting two from both sides of the equation, we have 𝑘 is equal to negative two. This is the value of 𝑘 we were trying to calculate.

In order to find the remaining critical numbers, we can now substitute 𝑘 equals negative two back into our quadratic function. This gives us zero is equal to negative 𝑥 squared plus two 𝑥 plus three 𝑥 minus four. Collecting like terms and then multiplying through by negative one, we have 𝑥 squared minus five 𝑥 plus four equals zero. Factoring our quadratic gives us 𝑥 minus one multiplied by 𝑥 minus four. And since one of these factors must equal zero, we have 𝑥 equals one and 𝑥 equals four. We already knew that there was a critical point at 𝑥 equals one. But we have now found that there is a third critical point at 𝑥 equals four.

We can therefore conclude that the function ℎ of 𝑥 equal to 𝑥 cubed plus 𝑘𝑥 squared multiplied by 𝑒 to the power of negative 𝑥 has critical points at 𝑥 equals zero, 𝑥 equals one, and 𝑥 equals four and the value of 𝑘 is negative two.