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 − 3x − 1 = 0

Accepted Answer

Welcome back, everyone. Apply the intermediate value theorem to confirm that the equation 3x cubed minus 7 X + 2 equals 0 has at least one solution specify the interval that contains the solution. We're given 4 answer choices A says from -4 to 2 inclusive, B from -1 to 0. Inclusive C from 0 to 1 inclusive, and D from 2 to 3 inclusive. Let's recall that the intermediate valid theorem can be applied for a continuous function in a specific interval and our function is a polynomial function on the left hand side of the equation, meaning it is continuous for any X, right? So essentially what we're going to do is take each of those intervals and test whether or not our function changes its sign at the end points of each interval. For example, at the beginning of the interval, it might be positive, and at the endpoint it changes sign to negative, right? Or It has a negative value and then changes assigned to positive. If that's the case, we can guarantee, according to the intermediate value theorem that there must be at least one root or 10, right? Meaning 3X minus 7 X + 2 equals 0 has at least one solution. So let's begin. Now for A, our endpoints are -4 and -2, so we want to evaluate F at -4. Which gives us 3 multiplied by -4 cubed minus 7 multiplied by -4 + 2. This gives us -162, and then we want to evaluate F at -2. So that's 3 multiplied by -2 cubed. -7 multiplied by -2 + 2, which gives us -8. Notice that we had a negative sign at the first endpoint and then we still have a negative sign at the other end point. So there is no change in sign, meaning A is not the correct option. We can exclude it. We can continue with B. We want to evaluate F at -1, which is 3 multiplied by -1 cubed minus 7 multiplied by -1 + 2. Which gives us 6, and then F adds 0. Which is 3 multiplied by 0 cubed minus 7 multiplied by 0 + 2, and that's 2. Notice that both values are positive, so there's no change in the sign, and that's not the interval that we're looking for. Let's move on to C. We want to valid F at 1 because we already know that F at 0 is equal to 2. So F of 1 is equal to 3 multiplied by 1 cubed minus 7 multiplied by 1 + 2. F of 1 gives us -2. So we had a negative sign, which turned into a positive sign. So this is the interval that we're looking for. We can see that there is, there is a change in sign. And therefore see the interval from 0 to an inclusive guarantees that there is a 0 in between, right, according to the intermediate value theorem, but we also might check option D to make sure that we can exclude that option. So here we can evaluate F of 2. Which is 3 multiplied by 2 cubed minus 7 multiplied by 2 + 2. F of 2 gives us 12, which is positive. And then F of 3. Which is 3 multiplied by 3 cubed. Minus 7 multiplied by 3 + 2. Which is 62, and it is still positive, so that's not the correct option, meaning the only correct answer is option C. The interval is from 0 to 1 inclusive. Thank you for watching.

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 − 3x − 1 = 0

Key Concepts

Intermediate Value Theorem

Continuous Functions

Graphical Analysis

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