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 | Nagwa 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 | Nagwa

# 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 • Second Year of Secondary School

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Let π(π₯) = 5 + ππ₯ + ππ₯Β². Suppose that the change in π(π₯) as π₯ goes from β1 to 2 is 6 and that the rate of change of π(π₯) at π₯ = 2 is 17. Determine π and π.

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### Video Transcript

Let π of π₯ be equal to five plus π times π₯ plus π times π₯ squared. Suppose that the change in π of π₯ as π₯ goes from negative one to two is six and that the rate of change of π of π₯ at π₯ is equal to two is 17. Determine π and π.

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

We recall that the change in a function π of π₯ from π₯ is equal to π to π₯ is equal to π is given by π evaluated at π minus π evaluated at π. Itβs the change in our function π as π₯ moves from π to π.

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

Finally, we have that five minus five is equal to zero. Two π plus π is equal to three π. And four π minus π is equal to three π. So by using the fact that the change in π of π₯ as π₯ went from negative one to two was six, weβve shown that six is equal to three π plus three π.

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

To do this, weβll use the second piece of information which weβre given. The rate of change of π of π₯ when π₯ is equal to two is 17. We recall the rate of change of π of π₯ when π₯ is equal to π is given by π prime of π is equal to the limit as β approaches zero of π evaluated at π plus β minus π evaluated at π all divided by β if this limit exists.

Since weβre told the rate of change of π of π₯ when π₯ is equal to two is 17, weβll set π equal to two. This gives us that 17 is equal to π prime evaluated at two. Using our definition for the rate of change, we have that this is equal to the limit as β approaches zero of π evaluated at two plus β minus π evaluated at two all divided by β.

Evaluating π at two plus β and two inside of our limit gives us the limit as β approaches zero of five plus π times two plus β plus π times two plus β squared minus five plus π times two plus π times two squared all divided by β. We see that weβre adding five and then subtracting five. So we can cancel these out.

Weβre now ready to start simplifying our limit. Weβll expand the first set of parentheses to get two π plus βπ. Next, we have that two plus β all squared is four plus four β plus β squared. Then we subtract two π and we subtract π times two squared, which is four π. Finally, we divide all of this by β.

We can simplify this further. We see that weβre adding two π and then subtracting two π. And we can also see that four times π is four π. And then we subtract four π. So we can cancel these out. Distributing π over our parentheses then gives us the limit as β approaches zero of β times π plus four βπ plus β squared π all divided by β.

We canβt yet evaluate this limit by using direct substitution. However, if we cancel the shared factor of β in our numerator and our denominator, we get the limit as β approaches zero of π plus four π plus βπ. One way of evaluating this is noticing this is a linear function in terms of β. So we can use direct substitution.

We substitute β is equal to zero to get π plus four π plus zero times π. And zero times π is just equal to zero. So we have the rate of change of our function π of π₯ when π₯ is equal to two is π plus four π. And weβre also told that this is equal to 17. So we now have two equations for π and π. We have that two is equal to π plus π and 17 is equal to π plus four π.

Thereβre several different ways of solving these simultaneous equations. Weβre going to subtract the bottom equation from the top equation. Subtracting 17 from two gives us negative 15. Subtracting π from π gives us zero. And then subtracting four π from π gives us negative three π. So we have that negative 15 is equal to negative three π. Dividing both sides of this equation by negative three gives us that π is equal to five.

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

Therefore, weβve shown if π of π₯ is equal to five plus ππ₯ plus π squared. And the change in π of π₯ as π₯ goes from negative one to two is six. And the rate of change of π of π₯ at π₯ is equal to two is 17. Then π is equal to negative three and π is equal to five.

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