Intermediate Value Theorem...3

[nycmathguy]

Section 2.2
Question 92

1. Let me start by saying that I don't have a graphing calculator to do part (b). If you have one, please answer part (b) in detail explaining everything concerning the adjustment of the table as the author instructs.

2. I will do 92 using the steps as shown by The Organic Chemistry Tutor on You Tube.

Question 92

h(x) = x^4 - 10x^2 + 3

I need to select a few x-values until a change occurs from negative positive or vice-versa.

How about 0 to start with?

When x = 0, h(0) = 3...positive.

How about 1?

h(1) = -6

This evaluation forms the interval
[0, 1].

There is a change in value from positive to negative. This means between h(0) and h(1) the y-value must cross zero. In other words, the y-value crosses the line x = 0. Using the Leading Coefficient Test applies here.

I will now find the real zeros in the interval [0, 1] by letting h(x) = 0 and solving for x.

0 = x^4 - 10x^2 + 3

Using Wolfram, the following values of x were found:

Question:

Which of the four values of x lies in our interval [0, 1]? To find out, I must use Wolfram.

The only value of x in our interval that yields zero is sqrt{5 - sqrt{22}}.

So, the answer is
x = sqrt{5 - sqrt{22}}.

Is any of this right?

 

[MathLover1]

it's correct for values of x lies in our interval [0, 1]

you need to check other intervals that might have zeros

use the graph to see where there is a change in value from positive to negative

 

[nycmathguy]

it's correct for values of x lies in our interval [0, 1]

you need to check other intervals that might have zeros

use the graph to see where there is a change in value from positive to negative

I will check other intervals on paper when time allows.