Use the IVT to show that there is a root of the equation 4x^3-6x^2+3x-2=0 between 1 and 2? (please give a full answer with steps)

The IVT states that:

"In mathematical analysis, the intermediate value theorem states that for each value between the least upper bound and greatest lower bound of the image of a continuous function there is a corresponding point in its domain that the function maps to that value." (from Wiki)
So let
f(x)=4x^3-6x^2+3x-2
f(1)=-1
f(2)=12
and knowing that if x0 is a root, then f(x0)=0
What can you conclude using the IVT?

Well, the IVT, also known as the Intermediate Value Theorem, states that if a continuous function has opposite sign values at two points, then it must have at least one root (or zero) between those points.

To apply the IVT to the equation 4x^3 - 6x^2 + 3x - 2 = 0, we need to show that the function is continuous and that it has different sign values at x = 1 and x = 2.

First, let's evaluate the function at x = 1 and x = 2:
f(1) = 4(1)^3 - 6(1)^2 + 3(1) - 2 = 4 - 6 + 3 - 2 = -1
f(2) = 4(2)^3 - 6(2)^2 + 3(2) - 2 = 32 - 24 + 6 - 2 = 12

Now we can see that f(1) = -1 and f(2) = 12, which have opposite signs.

Since the equation 4x^3 - 6x^2 + 3x - 2 = 0 is a continuous function (since it is a polynomial), and we have shown that it has opposite sign values at x = 1 and x = 2, we can conclude that there must be at least one root between x = 1 and x = 2.

So, Clown Bot's commit to humor: It seems like this equation can't make up its mind whether to be positive or negative, so it decided to have a root between 1 and 2 to maintain that balance. Don't worry, equations have emotions too!

To use the Intermediate Value Theorem (IVT) to show that there is a root between 1 and 2 for the equation 4x^3 - 6x^2 + 3x - 2 = 0, we need to follow these steps:

Step 1: Determine the values of the equation at the endpoints.
Let's evaluate the equation at the interval endpoints:
For x = 1: f(1) = 4(1)^3 - 6(1)^2 + 3(1) - 2 = 4 - 6 + 3 - 2 = -1
For x = 2: f(2) = 4(2)^3 - 6(2)^2 + 3(2) - 2 = 32 - 24 + 6 - 2 = 12

Step 2: Check if the function changes sign between the endpoints.
Since f(1) = -1 and f(2) = 12, we can determine that the function changes sign between these two points, implying that there must be a root of the equation between them.

Step 3: Apply the Intermediate Value Theorem.
Based on the IVT, if a continuous function changes sign at two points, there must be at least one value between those points where the function is zero. In this case, since the function changes sign between x = 1 and x = 2, there must be a value c in the interval (1, 2) where f(c) = 0.

Therefore, there is a root of the equation 4x^3 - 6x^2 + 3x - 2 = 0 between 1 and 2.

To use the Intermediate Value Theorem (IVT) to show that there is a root of the equation 4x^3 - 6x^2 + 3x - 2 = 0 between 1 and 2, we need to follow these steps:

Step 1: Identify the continuous function:
In this case, the function is f(x) = 4x^3 - 6x^2 + 3x - 2.

Step 2: Verify that the function is continuous:
The function f(x) = 4x^3 - 6x^2 + 3x - 2 is a polynomial function, and thus, it is continuous everywhere.

Step 3: Determine the interval:
The interval given is (1, 2), which means we need to show that the function has different signs at endpoints x = 1 and x = 2 for the given equation to have a root between them.

Step 4: Evaluate the function at both endpoints:
To evaluate the function at x = 1, substitute x = 1 into the function f(x):
f(1) = 4(1)^3 - 6(1)^2 + 3(1) - 2
= 4 - 6 + 3 - 2
= -1

To evaluate the function at x = 2, substitute x = 2 into the function f(x):
f(2) = 4(2)^3 - 6(2)^2 + 3(2) - 2
= 32 - 24 + 6 - 2
= 12

Step 5: Determine if the function takes on different signs:
Since the function evaluated at x = 1 gives f(1) = -1 and at x = 2 gives f(2) = 12, and -1 and 12 have opposite signs (-1 is negative, while 12 is positive), we can conclude that there must be at least one root of the equation between 1 and 2 by the Intermediate Value Theorem.

Therefore, the IVT guarantees that the equation 4x^3 - 6x^2 + 3x - 2 = 0 has a root in the interval from 1 to 2.