Video: Finding Unknown Coefficients in a Polynomial Function given the Value of the Function and Its Derivative at a Point

Suppose that, for function 𝑓(π‘₯) = π‘Žπ‘₯Β³ + 𝑏π‘₯, we have 𝑓(βˆ’1) = 12 and 𝑓′(βˆ’1) = βˆ’20. Determine π‘Ž and 𝑏.

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Video Transcript

Suppose that, for function 𝑓 of π‘₯ is equal to π‘Žπ‘₯ cubed plus 𝑏π‘₯, we have 𝑓 of negative one is equal to 12 and 𝑓 prime of negative one is equal to negative 20. Determine π‘Ž and 𝑏.

We’re given a cubic function 𝑓 of π‘₯. And we’re told that 𝑓 evaluated at negative one is equal to 12 and 𝑓 prime evaluated at negative one is equal to negative 20. We need to use this to determine the values of π‘Ž and 𝑏. First, we know we can find 𝑓 evaluated at negative one by substituting π‘₯ is equal to negative one into our function 𝑓 of π‘₯. But we also need to remember what 𝑓 prime evaluated at negative one means.

We recall since 𝑓 is a function of π‘₯, 𝑓 prime of π‘₯ means the derivative of 𝑓 with respect to π‘₯. And we need to evaluate this when π‘₯ is equal to negative one. Luckily, our function 𝑓 of π‘₯ is a cubic polynomial. And we know how to differentiate this term by term by using the power rule for differentiation. So, let’s start by finding an expression for 𝑓 prime of π‘₯. That’s the derivative of π‘Žπ‘₯ cubed plus 𝑏π‘₯ with respect to π‘₯. And since this is a polynomial, we’ll differentiate this term by term by using the power rule for differentiation.

We recall this tells us for any constants 𝑐 and 𝑛, the derivative of 𝑐π‘₯ to the 𝑛th power with respect to π‘₯ is equal to 𝑛 times 𝑐π‘₯ to the power of 𝑛 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‘Ž 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‘Ž times π‘₯ squared.

We now want to differentiate our second term with respect to π‘₯. We could do this by writing it as 𝑏 times π‘₯ to the first power. We could then apply our power rule for differentiation with our value of 𝑛 equal to one. However, an alternative method for doing this is remembering that 𝑏π‘₯ 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 𝑏. 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 𝑓 prime of π‘₯ is equal to three π‘Žπ‘₯ squared plus 𝑏.

We’re now ready to use both of our pieces of information about 𝑓 of π‘₯. 𝑓 prime of negative one is equal to negative 20 and 𝑓 of negative one is equal to 12. Let’s start with 𝑓 evaluated at negative one. We need to substitute π‘₯ is equal to negative one into our function 𝑓 of π‘₯. We get π‘Ž times negative one all cubed plus 𝑏 times negative one. We can simplify the right-hand side of this equation to get negative π‘Ž minus 𝑏. And remember, we’re told that 𝑓 evaluated at negative one is equal to 12. So, we can just equate this equal to 12.

We can do the same to find 𝑓 prime evaluated at negative one. We’d substitute π‘₯ is equal to negative one into our expression for 𝑓 prime of π‘₯. We get three π‘Ž times negative one squared plus 𝑏. And we can simplify the right-hand side of this equation to give us three π‘Ž plus 𝑏. And remember, 𝑓 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‘Ž and 𝑏. So, we need to solve this pair of simultaneous equations to find our values of π‘Ž and 𝑏.

There’re several different ways of doing this. For example, we could rearrange one of our equations to make π‘Ž or 𝑏 the subject. However, we’ll do this by adding our two equations together. This is because one equation contains negative 𝑏 and the other one contains 𝑏. So, adding these two equations together will eliminate our variable 𝑏. Let’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‘Ž to three π‘Ž, we get two π‘Ž. Finally, negative 𝑏 plus 𝑏 is equal to zero. So, we have negative eight is equal to two π‘Ž. We can solve for π‘Ž by dividing through by two. And, of course, negative eight divided by two is negative four. So, we’ve shown that π‘Ž is equal to negative four. And we can then find the value of 𝑏 by substituting π‘Ž is equal to negative four into one of our equations.

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

Therefore, if the function 𝑓 of π‘₯ is equal to π‘Žπ‘₯ cubed plus 𝑏π‘₯ has 𝑓 evaluated at negative one is equal to 12 and 𝑓 prime evaluated at negative one is equal to negative 20, we’ve shown that π‘Ž must be equal to negative four and 𝑏 must be equal to negative eight.

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