From the given information:
We are to consider the following equation:
[tex]cos (x) = x^3[/tex]
(a)
To prove that the equation has at least one real root;
The equation can be expressed as:
[tex]\mathbf{cos (x) = x^3}[/tex]
Given that the function of (x) is;
i.e.
[tex]\mathbf{f(x) = cos(x) - x^3}[/tex] is continuous on the interval [0, 1]:
Assuming we replace x with 0 in the above equation;
Then,
f(0) = cos (0) - (0)³
f(0) = 1 - 0
f(0) = 1
∴
So, f(0) = Cos (0) - (0)³= 1 > 0
Similarly, let's replace x with 1;
Then,
f(1) = cos (1) - (1)³
f(1) = -0.459698
f(1) ≅ -0.46
∴
So, f(1) = Cos (1) - (1)³= -0.46 < 0
By applying these two results, we have;
1 > 0 > -0.46 i.e. f(0) > 0 > f(1)
Since 1 > 0 > -0.46, by intermediate theorem, there is a number c in (0, 1) such that f(c) = 0
Thus, the root of the equation cos x - x³ = 0 in the interval (0, 1).
(b)
Taking the equation cos(x) = x³ into account by using a graphing calculator.
The graphing calculator's trace feature finds the intersection point for the two functions cos(x) and x³, as seen in the figure below.
We will notice from the graph that only one root of the equation for cos(x) = x³ is;
x = 0.865474
x ≅ 0.86
∴
As a result, the needed interval length of 0.01 is as follows:
x ∈ (0.86, 0.87)
Therefore, we can conclude that the equation is being proved from the above equation and the interval length is determined by using a calculator.
Learn more about graphing calculator here:
https://brainly.com/question/9339041?referrer=searchResults
