Consider the following. cos(x) = x^3 (a) Prove that the equation has at least one real root. The equation cos(x) = x^3 is equivalent to the equation f(x) = cos(x) − x^3 = 0. f(x) is continuous on the interval [0, 1], f(0) = , and f(1) = . Since < 0 < , there is a number c in (0, 1) such that f(c) = 0 by the Intermediate Value Theorem. Thus, there is a root of the equation cos(x) = x^3, in the interval (0, 1). (b) Use your calculator to find an interval of length 0.01 that contains a root. (Enter your answer using interval notation. Round your answers to two decimal places.)

If f is a continuous function on the interval [a,b], then f assumes any value between f(a) and f(b) at some point within the interval, i.e., for every z in the image of f, there is w in [a,b] such that z = f(w).

An important consequence of this result is that if the continuous function f has values of opposite sign inside the interval [a,b], then f has a root in that interval, i.e., there is w in [a,b] satisfying f(w)=0.

We are going to show that the function f defined by[tex]f(x)=\cos(x)-x^3[/tex] has a real root by seeing that f changes its sign on the interval [0, 1]. In fact:

[tex]f(0)=\cos(0)-0^3=0[/tex]

[tex]f(1)=\cos(1)-1^3\approx -0.46[/tex]

Then, as a consequence of the IVT, there is w in [0,1] such that f(w)=0, or equivalently, [tex]\cos(w)=w^3[/tex], as desired.

Now, doing some calculations we found that the interval [tex]I=[0.86,0.87][/tex] contains the root w of f, since f changes its sign here:

[tex]f(0.86)=cos(0.86)-(0.86)^3\approx 0.02[/tex]

[tex]f(0.87)=cos(0.87)-(0.87)^3\approx -0.01[/tex]

Furthermore, the length of the interval I is: [tex]0.87-0.86=0.01[/tex]. (See the graph)