Roots (Zeros)

Show that the ...

Question

Roots (Zeros)

Show that the functions in Exercises 19–26 have exactly one zero in the given interval.

f(x) = x⁴ + 3x + 1, [−2, −1]

Accepted Answer

Hello. In this video, we are going to be verifying that the given function H of X has exactly 10 on the interval from -3 to -1.5. In order to check to see if this function has 10, we are going to use the intermediate value theorem. The intermediate value theorem has two main conditions we need to check. First, we need to check to see that the function. Is continuous. On our interval. And if the function is continuous on the interval, then we need to check to see if the end points or the values of the endpoints are of opposite signs. So, for example, if we plug in H of -3, and it is less than H of -1.5, then the function has to cross the X axis at least once, or the opposite can be true. We want to show that H of -1.5 is less than H of -3. So, we will go ahead and start by checking the first condition. We need to ask ourselves, is the given function H of X continuous on our interval? So, H of X is given to us as 2 X 5th power minus 5 X cubed plus X + 2. Since this function is in the form of a polynomial, then the polynomial is defined on all values of X. So the domain of the function is from negative infinity to positive infinity. Thus, it is going to be continuous from -3 to -1.5. So the function is continuous on the interval. Now we need to check the values of the end of the endpoints of the function. So for the second part of this function, or for this problem, we are going to plug in the end points into our function. We will start with -3. H of -3 will change the function to be 2 multiplied by -3, raise to the 5th power, minus 5 multiplied by -3, raise to the third power, -3 + 2. Simplifying this function, -3 to 5th power, multiplied by 2, will give us the value of -486. -5, multiplied by -3, rates to the third power, will give us the value of 135. Then we have -3 + 2. And taking the sum of all these values together will give us a final sum of negative 352. So, what this means is that when we plug in the boundary 0.-3 into our function, we get a value that is less than 0. So currently we have H of -3 is less than 0. Now we need to check the value of the boundary point at -1.5. So, we are going to plug in -1.5 into our function. That will be H of -1.5 is equal to 2 multiplied by -1.5, raised to the fifth power, minus 5 multiplied by -1.5, raise to the third power, minus 1.5 plus 2. Two, multiplied by -1.5, rate to the fifth power will be negative 15.19. When we raise around to the nearest 100th place. -5, multiplied by -1.5, raised to the third power, will simplify to be 16.88. Then we have 1.5 plus 2, and summing these values together is going to give us a value of 2.19, which is a positive value greater than 0. So what we have here is H of -1.5. Which is a positive value greater than 0. And what this means is that H of -3 is going to be less than H of -1.5, and that means that this graph or this polynomial, however it may look, is going to pass through the X axis at least once, meaning that there is going to be at least 10 for the function. So the answer to this problem is yes, there is going to be 10 for the given function. So I hope this video helps you in understanding on how to approach this problem, and I will go ahead and see you all in the next video.

Roots (Zeros)

Show that the functions in Exercises 19–26 have exactly one zero in the given interval.

f(x) = x⁴ + 3x + 1, [−2, −1]

Key Concepts

Intermediate Value Theorem

Continuity of Polynomial Functions

Derivative and Monotonicity

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