Use the Intermediate Value Theorem in Exercises 69–74 to prove...

Question

Use the Intermediate Value Theorem in Exercises 69–74 to prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations.

x³ − 15x + 1 = 0 (three roots)

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says to apply the intermediate value theorem to confirm that the equation XQ + 3X2 minus 4 equal to 0 has at least one solution. Specify the interval that contains this solution. We give 4 possible choices as our answers. For choice A, we have the open interval from 0 to 2. For choice B, we have the open interval from -1 to 0. For choice C, we have the open interval from -4 to -3. And for choice D, we have the open interval from 0 to 1/2. Now, we need to apply the intermediate value theorem to confirm that the equation XU plus 3X2 minus 4 equal to 0 has at least one solution. So we're gonna need to evaluate the left-hand side of that equation for a couple of values of X. So we're gonna start off with X equal to 0. If we call the left-hand side of the equation given in the problem as F of X, then we'll evaluate F of 0, which is equal to 0 cubed, plus 3 multiplied by 0 squared minus 4. And when we evaluate this expression, this is going to be equal to -4. The next point we'll look at is when X is equal to 2. And so we'll evaluate F of 2, which is equal to 2 cubed, plus 3 multiplied by 2 squared minus 4. And evaluating this expression, this is equal to 16. Now, since F of 0 is less than 0 and F of 2 is greater than 0, then by the intermediate value theorem, there is at least one root within this interval. And so, that means that the answer to this problem that we're looking for is the open interval from 0 to 2. Since the solution is contained within that interval. And if we look at our answer choices, the correct answer choice actually corresponds to answer choice A. If we look at the remaining answer choices, if we evaluate our function F of X at the end points of those intervals, our function F of X always evaluates to a negative value. And so, therefore, the intermediate value theorem does not guarantee that there is at least one solution in that interval. So I hope that this explanation has been useful, and I'll see you in the next video.

Use the Intermediate Value Theorem in Exercises 69–74 to prove that each equation has a solution. Then use a graphing calculator or computer grapher to solve the equations.
x³ − 15x + 1 = 0 (three roots)

Key Concepts

Intermediate Value Theorem

Continuous Functions

Graphing Techniques

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