[Technology Exercise] Roots

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Question

[Technology Exercise] Roots

Let ƒ(𝓍) = 𝓍³ ―𝓍― 1.

a. Use the Intermediate Value Theorem to show that ƒ has a zero between ―1 and 2 .

Accepted Answer

Welcome back everyone. Consider the function H X equals 2 x cubed minus 3X2 minus 12 x + 1. Using the intermediate value theorem, identify the interval that must contain a root of H. We're given 4 answer choices A says from -2 to 1 inclusive, B from 2 to 3 inclusive, C from 2 to -1 inclusive, and D from 1 to 2 inclusive. So, let's recall that the intermediate value theorem says that if we are considering a specific closed interval in the form of A to B, Those values are inclusive, right, if our h of A is positive while H of B is negative or vice versa, such as H of A is negative and H of B. is positive. This guarantees that we have at least one route for H of X in the given interval between A to B, right? Simply speaking, we want to make sure that at the end points our age of X changes sign. So the way we're going to use in this problem to solve it is simply by evaluating H of X at each endpoint. So we can begin with the answer choice A. Our initial point is -2, so we want to validate H at -2. And this means that we're taking 2 multiplied by -2 cubed, minus 3 multiplied by -2 squad, minus 12 multiplied by -2 + 1. And what we're going to do is simply perform the calculation. This gives us -3, and we understand that this is negative. So according to the intermediate value theorem, if we evaluate H at the other end point, which is 1, we want to make sure that we're getting a positive value. This guarantees that there's at least one route, right? So let's evaluate H at 1, which means that we're taking 2 multiplied by 1 cubed minus 3 multiplied by 1 squared minus 12 multiplied by 1 + 1. And this gives us -12. It is still negative. We did not observe a change in sign, meaning A is not the correct option, right? We essentially want to change the sign of H when we are considering the second endpoint, and this is not the case because both of these are negative values. So we're going to approach the other options using the same way. First of all, let's evaluate H at. 2 for option B. If we perform the same calculations. We're going to get H of 2 equals -19, which is negative, and each at the other end point, which is 3. Is going to be equal to -8, which is also negative, so there is no change in sign, right, meaning B is not the interval that we're looking for. For option C, we went to valid H at -2. We already know that it is equal to -3. It is negative. And now we want to valid H at -1. Now if we validate H at -1, we're going to get 8, which is positive. So this is where we notice a sign change from negative to positive, meaning C is the correct option so far. And now let's consider option D. We want to valid H at 1. We already know that it's -12. Which is negative and then we want to evaluate H at 0.2. We already know that it's -19, so those two are negative, and that means it is not the correct option. Essentially, the interval that must contain a root of H is the interval C because in this interval, The value of each at the end points changes sign, right? And that's our final answer. Thank you for watching.

[Technology Exercise] Roots

Let ƒ(𝓍) = 𝓍³ ―𝓍― 1.

a. Use the Intermediate Value Theorem to show that ƒ has a zero between ―1 and 2 .

Watch next

[Technology Exercise] RootsLet ƒ(𝓍) = 𝓍³ ―𝓍― 1.a. Use the Intermediate Value Theorem to show that ƒ has a zero between ―1 and 2 .

Watch next

[Technology Exercise] Roots

Let ƒ(𝓍) = 𝓍³ ―𝓍― 1.

a. Use the Intermediate Value Theorem to show that ƒ has a zero between ―1 and 2 .