# How do I show that the equation

x^4 + 3x + 1 = 0, -2 <= x <= -1

has exactly one solution in the interval.

Thanks.

One way to do this is to use trial and error. split the interval (-2,-1) into 10 equal parts. Then evaluate the function at each point. That is put the value of x as -2, -1.9,-1.8......-1. Look for changes in sign. That is the value of function should change signs, if this happens only once, then the function has only one solution in the interval. The values of x^4 + 3x + 1 at -2, -1.9, -1.8.......-1 are 11, 8.33,6.09, 4.25,2.75,1.56,0.6416,-0.0439,-0.53,-0.84,-1. Its very clear that the sign change occurs only once when X transitions from -1.4 to -1.3 so there is only one solution in this range which is between -1.4 and -1.3. This is however not an exact method.

The exact method would be to show first that the function is monotoniaclly increasing or monotonically decreasing and then show that the function takes a value of 0 in this interval. To do this you need to use calculus.

Yes, calculus is what I want to use!

This is from the chapter "application of derivatives".

Using mean value theorem how can i explain this? Thanks.

## To prove that the equation x^4 + 3x + 1 = 0 has exactly one solution in the interval -2 ≤ x ≤ -1, you can use the Mean Value Theorem from calculus. Here's how you can explain it:

1. Start by showing that the function f(x) = x^4 + 3x + 1 is continuous on the interval [-2, -1]. This means that the function has no holes, jumps, or vertical asymptotes in this interval.

2. Next, show that f(x) is differentiable on (-2, -1), which means it has a derivative at every point within this interval. In this case, the derivative of f(x) is f'(x) = 4x^3 + 3.

3. Apply the Mean Value Theorem, which states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c such that f'(c) = (f(b) - f(a))/(b - a).

4. In this case, a = -2, b = -1, and f(x) = x^4 + 3x + 1. So, we have f(-2) = (-2)^4 + 3(-2) + 1 = 23 and f(-1) = (-1)^4 + 3(-1) + 1 = -1.

5. Therefore, the Mean Value Theorem guarantees the existence of a number c in the interval (-2, -1) such that f'(c) = (23 - (-1))/(-2 - (-1)) = 24.

6. Solve the equation f'(c) = 24, which is 4c^3 + 3 = 24. Simplifying, we get 4c^3 = 21, and further solving yields c ≈ -1.385.

7. Since c is the solution to f'(c) = 24, it implies that f(c) = 0, because the derivative of f(x) represents the rate of change of the function, and when it equals 24, the function is increasing and crosses the x-axis.

8. Therefore, we have shown that there exists at least one solution to x^4 + 3x + 1 = 0 in the interval -2 ≤ x ≤ -1, and since f(c) = 0 is the only solution, there is exactly one solution in this interval.

By applying the Mean Value Theorem and considering the properties of the function and its derivative, you have mathematically proven that the equation has exactly one solution in the given interval.