Question Video: Finding Unknowns in a Polynomial Function given Its Rate of Change at a Point and Its Average Rate of Change at Two Points Mathematics • Higher Education

Let 𝑓(π‘₯) = 5 + π‘Žπ‘₯ + 𝑏π‘₯Β². Suppose that the change in 𝑓(π‘₯) as π‘₯ goes from βˆ’1 to 2 is 6 and that the rate of change of 𝑓(π‘₯) at π‘₯ = 2 is 17. Determine π‘Ž and 𝑏.

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Video Transcript

Let 𝑓 of π‘₯ be equal to five plus π‘Ž times π‘₯ plus 𝑏 times π‘₯ squared. Suppose that the change in 𝑓 of π‘₯ as π‘₯ goes from negative one to two is six and that the rate of change of 𝑓 of π‘₯ at π‘₯ is equal to two is 17. Determine π‘Ž and 𝑏.

The question gives us a polynomial function 𝑓 of π‘₯ with two unknown constant variables π‘Ž and 𝑏. The question wants us to determine the values of π‘Ž and 𝑏 using the following information. The change in 𝑓 of π‘₯ as π‘₯ goes from negative one to two is six. And the rate of change of 𝑓 of π‘₯ at π‘₯ is equal to two is 17.

We recall that the change in a function 𝑔 of π‘₯ from π‘₯ is equal to 𝑐 to π‘₯ is equal to 𝑑 is given by 𝑔 evaluated at 𝑑 minus 𝑔 evaluated at 𝑐. It’s the change in our function 𝑔 as π‘₯ moves from 𝑐 to 𝑑.

The question tells us the change in 𝑓 of π‘₯ as π‘₯ goes from negative one to two is six. In other words, 𝑓 evaluated at two minus 𝑓 evaluated at negative one is equal to six. Substituting π‘₯ is equal to two and π‘₯ is equal to negative one into our function 𝑓 of π‘₯ gives us that the change in 𝑓 of π‘₯ as π‘₯ goes from negative one to two is equal to five plus π‘Ž times two plus 𝑏 times two squared minus five plus π‘Ž times negative one plus 𝑏 times negative one squared. We can evaluate this to give us five plus two π‘Ž plus four 𝑏 minus five plus π‘Ž minus 𝑏.

Finally, we have that five minus five is equal to zero. Two π‘Ž plus π‘Ž is equal to three π‘Ž. And four 𝑏 minus 𝑏 is equal to three 𝑏. So by using the fact that the change in 𝑓 of π‘₯ as π‘₯ went from negative one to two was six, we’ve shown that six is equal to three π‘Ž plus three 𝑏.

In fact, we can divide both sides of this equation by three to get that two is equal to π‘Ž plus 𝑏. Since π‘Ž and 𝑏 can be any constants, the equation two is equal to π‘Ž plus 𝑏 has many solutions. So we need to get more information to find the values of π‘Ž and 𝑏.

To do this, we’ll use the second piece of information which we’re given. The rate of change of 𝑓 of π‘₯ when π‘₯ is equal to two is 17. We recall the rate of change of 𝑓 of π‘₯ when π‘₯ is equal to π‘Ž is given by 𝑓 prime of π‘Ž is equal to the limit as β„Ž approaches zero of 𝑓 evaluated at π‘Ž plus β„Ž minus 𝑓 evaluated at π‘Ž all divided by β„Ž if this limit exists.

Since we’re told the rate of change of 𝑓 of π‘₯ when π‘₯ is equal to two is 17, we’ll set π‘Ž equal to two. This gives us that 17 is equal to 𝑓 prime evaluated at two. Using our definition for the rate of change, we have that this is equal to the limit as β„Ž approaches zero of 𝑓 evaluated at two plus β„Ž minus 𝑓 evaluated at two all divided by β„Ž.

Evaluating 𝑓 at two plus β„Ž and two inside of our limit gives us the limit as β„Ž approaches zero of five plus π‘Ž times two plus β„Ž plus 𝑏 times two plus β„Ž squared minus five plus π‘Ž times two plus 𝑏 times two squared all divided by β„Ž. We see that we’re adding five and then subtracting five. So we can cancel these out.

We’re now ready to start simplifying our limit. We’ll expand the first set of parentheses to get two π‘Ž plus β„Žπ‘Ž. Next, we have that two plus β„Ž all squared is four plus four β„Ž plus β„Ž squared. Then we subtract two π‘Ž and we subtract 𝑏 times two squared, which is four 𝑏. Finally, we divide all of this by β„Ž.

We can simplify this further. We see that we’re adding two π‘Ž and then subtracting two π‘Ž. And we can also see that four times 𝑏 is four 𝑏. And then we subtract four 𝑏. So we can cancel these out. Distributing 𝑏 over our parentheses then gives us the limit as β„Ž approaches zero of β„Ž times π‘Ž plus four β„Žπ‘ plus β„Ž squared 𝑏 all divided by β„Ž.

We can’t yet evaluate this limit by using direct substitution. However, if we cancel the shared factor of β„Ž in our numerator and our denominator, we get the limit as β„Ž approaches zero of π‘Ž plus four 𝑏 plus β„Žπ‘. One way of evaluating this is noticing this is a linear function in terms of β„Ž. So we can use direct substitution.

We substitute β„Ž is equal to zero to get π‘Ž plus four 𝑏 plus zero times 𝑏. And zero times 𝑏 is just equal to zero. So we have the rate of change of our function 𝑓 of π‘₯ when π‘₯ is equal to two is π‘Ž plus four 𝑏. And we’re also told that this is equal to 17. So we now have two equations for π‘Ž and 𝑏. We have that two is equal to π‘Ž plus 𝑏 and 17 is equal to π‘Ž plus four 𝑏.

There’re several different ways of solving these simultaneous equations. We’re going to subtract the bottom equation from the top equation. Subtracting 17 from two gives us negative 15. Subtracting π‘Ž from π‘Ž gives us zero. And then subtracting four 𝑏 from 𝑏 gives us negative three 𝑏. So we have that negative 15 is equal to negative three 𝑏. Dividing both sides of this equation by negative three gives us that 𝑏 is equal to five.

We can then find the value of π‘Ž by substituting 𝑏 is equal to five into either of our simultaneous equations. Substituting 𝑏 is equal to five into the equation two is equal to π‘Ž plus 𝑏 gives us two is equal to π‘Ž plus five. And subtracting five from both sides of this equation gives us that π‘Ž is equal to negative three.

Therefore, we’ve shown if 𝑓 of π‘₯ is equal to five plus π‘Žπ‘₯ plus 𝑏 squared. And the change in 𝑓 of π‘₯ as π‘₯ goes from negative one to two is six. And the rate of change of 𝑓 of π‘₯ at π‘₯ is equal to two is 17. Then π‘Ž is equal to negative three and 𝑏 is equal to five.

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