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) = x³ + 4x² + 7, (−∞, 0)

Accepted Answer

Hello. In this video, we are going to be verifying that the given function JFX equal to 3X cubed plus 9X2 minus 30 X on the interval from negative infinity to -2, has exactly 10. In order to verify this, we are going to use the intermediate value theorem. Now there are 2 main conditions that we need to check for the intermediate value theorem. The first condition is to check that the function is continuous on our interval. And the second condition is we want to show that the end points J of A is less than 0. And J of B is less than 0, or greater than 0. Or we want to show that the other way around, J of B is less than 0, and J A is greater than 0. So, we will go ahead and start by checking for continuity of the function. Now we want to check to make sure that our function is going to be continuous from negative affinity to -2. Since the function given to us is in the form of a polynomial, its domain is going to be all real numbers. Since the domain is all real numbers, that means that the function will be continuous from the interval from negative infinity to -2. So the first condition is met for the intermediate value theorem. Now we need to check the 2nd condition for the intermediate value theorem. In order to check this condition, we are going to be plugging in the boundaries of our interval into the function. So we will go ahead and start by plugging in negative infinity into our JFX function. If we plug in negative infinity, we get 3 multiplied by negative infinity cubed, plus 9 multiplied by negative infinity squared, minus 30 multiplied by negative infinity. Now, the thing about this problem is that we want to take a look at the leading term of this polynomial. Negative infinity cubed will simplified to be negative infinity, and anything multiplied to negative infinity will give us negative infinity. Since negative infinity is the leading term in this polynomial, that means the value of J with negative infinity plugged into it is going to be negative infinity, and this is going to be a value less than 0. So, what we have right now is that J of negative infinity is less than 0. Now we want to check to see if J-2 will be greater than 0. So, when we plug in -2 into our function, we get J of -2 is equal to 3 multiplied by -2 cubed, plus 9 multiplied by -2 square minus 30 multiplied by -2. 3, multiplied by -2 cubed is going to give us the value of -24. 9 multiplied by -2 squad will simplified to give us 36. And -30 multiplied by -2, will give us 60. And summing these three values together will give us a final value of 72, which is greater than 0. What we have here is that J-2 is greater than 0, and what this means is that on our interval from negative infinity to -2, since the function is continuous and the end points are opposite signs negative and positive. This graph is going to have to cross the X axis at least once, meaning that there is going to be at least 10 for the given function, and that is going to be the solution to this problem by the intermediate value theorem. 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) = x³ + 4x² + 7, (−∞, 0)

Key Concepts

Intermediate Value Theorem

Continuity of Polynomial Functions

Behavior of Polynomial Functions at Infinity

Watch next