# Question Video: Finding Unknown Coefficients in a Polynomial Function given the Value of the Function and Its Derivative at a Point Mathematics • Higher Education

Suppose that, for function π(π₯) = ππ₯Β³ + ππ₯, we have π(β1) = 12 and πβ²(β1) = β20. Determine π and π.

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### Video Transcript

Suppose that, for function π of π₯ is equal to ππ₯ cubed plus ππ₯, we have π of negative one is equal to 12 and π prime of negative one is equal to negative 20. Determine π and π.

Weβre given a cubic function π of π₯. And weβre told that π evaluated at negative one is equal to 12 and π prime evaluated at negative one is equal to negative 20. We need to use this to determine the values of π and π. First, we know we can find π evaluated at negative one by substituting π₯ is equal to negative one into our function π of π₯. But we also need to remember what π prime evaluated at negative one means.

We recall since π is a function of π₯, π prime of π₯ means the derivative of π with respect to π₯. And we need to evaluate this when π₯ is equal to negative one. Luckily, our function π of π₯ is a cubic polynomial. And we know how to differentiate this term by term by using the power rule for differentiation. So, letβs start by finding an expression for π prime of π₯. Thatβs the derivative of ππ₯ cubed plus ππ₯ with respect to π₯. And since this is a polynomial, weβll differentiate this term by term by using the power rule for differentiation.

We recall this tells us for any constants π and π, the derivative of ππ₯ to the πth power with respect to π₯ is equal to π times ππ₯ to the power of π minus one. We multiply by our exponent of π₯ and then reduce this exponent by one. In our first term, the exponent of π₯ is equal to three. So we multiply by our exponent of three and then reduce this exponent by one. This gives us three π times π₯ to the power of three minus one. And of course, we can simplify three minus one to give us two. So, the derivative of our first term with respect to π₯ was three π times π₯ squared.

We now want to differentiate our second term with respect to π₯. We could do this by writing it as π times π₯ to the first power. We could then apply our power rule for differentiation with our value of π equal to one. However, an alternative method for doing this is remembering that ππ₯ is a linear function. And the slope of a linear function will just be the coefficient of π₯, which in this case is equal to π. And this will always work. The derivative of a constant multiplied by π₯ with respect to π₯ will always just be the coefficient of π₯. So, we have π prime of π₯ is equal to three ππ₯ squared plus π.

Weβre now ready to use both of our pieces of information about π of π₯. π prime of negative one is equal to negative 20 and π of negative one is equal to 12. Letβs start with π evaluated at negative one. We need to substitute π₯ is equal to negative one into our function π of π₯. We get π times negative one all cubed plus π times negative one. We can simplify the right-hand side of this equation to get negative π minus π. And remember, weβre told that π evaluated at negative one is equal to 12. So, we can just equate this equal to 12.

We can do the same to find π prime evaluated at negative one. Weβd substitute π₯ is equal to negative one into our expression for π prime of π₯. We get three π times negative one squared plus π. And we can simplify the right-hand side of this equation to give us three π plus π. And remember, π prime of negative one is equal to negative 20. So, we can just equate this equal to negative 20. We now have a pair of simultaneous equations in π and π. So, we need to solve this pair of simultaneous equations to find our values of π and π.

Thereβre several different ways of doing this. For example, we could rearrange one of our equations to make π or π the subject. However, weβll do this by adding our two equations together. This is because one equation contains negative π and the other one contains π. So, adding these two equations together will eliminate our variable π. Letβs start by adding the left-hand sides of the equations together. We get 12 plus negative 20, which we know is equal to negative eight.

Next, when we add negative π to three π, we get two π. Finally, negative π plus π is equal to zero. So, we have negative eight is equal to two π. We can solve for π by dividing through by two. And, of course, negative eight divided by two is negative four. So, weβve shown that π is equal to negative four. And we can then find the value of π by substituting π is equal to negative four into one of our equations.

Substituting π is equal to negative four into our first equation gives us 12 is equal to negative one times negative four minus π. And then, we can just rearrange and solve this equation. If we do this, we find that π is equal to negative eight.

Therefore, if the function π of π₯ is equal to ππ₯ cubed plus ππ₯ has π evaluated at negative one is equal to 12 and π prime evaluated at negative one is equal to negative 20, weβve shown that π must be equal to negative four and π must be equal to negative eight.