show that the equation xcosx-2x2+3x-1=0 has at least one solution in [0.2,0.3]

xcosx-2x2+3x-1=0

Evaluate the function at 0.2 and 0.3:

0.2*cos0.2-2*0.2^2+3*0.2-1 = -.280

0.3*cos0.3-2*0.3^2+3*0.3-1 = 0.019

In order for the function to go from y=-.280 to y=0.019 on the interval [0.2,0.3], it must pass through y = 0 at least once.

-x*sinx -4x + 3 =0

To show that the equation `x*cos(x) - 2*x^2 + 3*x - 1 = 0` has at least one solution in the interval `[0.2, 0.3]`, we can use the Intermediate Value Theorem.

The Intermediate Value Theorem states that if a function is continuous on a closed interval `[a, b]` and takes on values f(a) and f(b) on the interval, then it must also take on every value between f(a) and f(b) at some point within the interval.

In this case, let's define a new function `f(x)` such that `f(x) = x*cos(x) - 2*x^2 + 3*x - 1`.

To apply the Intermediate Value Theorem, we need to first verify that `f(x)` is continuous on the interval `[0.2, 0.3]`.

Since `f(x)` is a combination of continuous functions, namely `x`, `cos(x)`, and polynomials, it is also continuous over that interval.

Now, we can evaluate `f(0.2)` and `f(0.3)` to check if `f(x)` takes on different signs at the endpoints of the interval:

`f(0.2) = 0.2*cos(0.2) - 2*(0.2)^2 + 3*(0.2) - 1 ≈ -0.732`
`f(0.3) = 0.3*cos(0.3) - 2*(0.3)^2 + 3*(0.3) - 1 ≈ 0.028`

Since `f(0.2)` is negative and `f(0.3)` is positive, we can conclude that `f(x)` takes on both positive and negative values within the interval `[0.2, 0.3]`. Therefore, according to the Intermediate Value Theorem, there must be at least one solution to the equation `x*cos(x) - 2*x^2 + 3*x - 1 = 0` in the interval `[0.2, 0.3]`.

To show that the equation \(x\cos(x) - 2x^2 + 3x - 1 = 0\) has at least one solution in the interval \([0.2, 0.3]\), we can apply the Intermediate Value Theorem. According to this theorem, if a continuous function takes on values of opposite signs at the endpoints of an interval, it must pass through zero (or take on the value zero) at least once within that interval.

Here's how we can apply the Intermediate Value Theorem step-by-step to prove that the given equation has a solution in the interval \([0.2, 0.3]\):

Step 1: Evaluate the endpoints of the interval
We need to check the values of the equation at both ends of the interval. Substitute \(x = 0.2\) and \(x = 0.3\) into the equation:

For \(x = 0.2\):
\(0.2\cos(0.2) - 2(0.2)^2 + 3(0.2) - 1 \approx -0.650\)

For \(x = 0.3\):
\(0.3\cos(0.3) - 2(0.3)^2 + 3(0.3) - 1 \approx 0.150\)

Step 2: Check for opposite signs
Since the value at \(x = 0.2\) is negative (-0.650) and the value at \(x = 0.3\) is positive (0.150), we can conclude the function has values of opposite signs at the endpoints of the interval.

Step 3: Conclude by the Intermediate Value Theorem
By the Intermediate Value Theorem, since the function takes on values of opposite signs at the endpoints of the interval \([0.2, 0.3]\), we can be certain that it must have at least one zero (solution) somewhere within that interval.

Therefore, we have shown that the equation \(x\cos(x) - 2x^2 + 3x - 1 = 0\) has at least one solution in the interval \([0.2, 0.3]\).