Consider the following. cos(x) = x3
A) Prove that the equation has at least one real root.
The equation cos(x) = x3 is equivalent to the equation f(x) = cos(x) − x3 = 0. f(x) is continuous on the interval (0, 1) 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) = x3, in the interval (0, 1).
B) Use your calculator to find an interval of length 0.01 that contains a root.

Step-by-step explanation:

Given that:

cos(x) = x³

f(x) = cos (x) - x³

f(x) is continuous on the interval (0, 1)

when;

f(0) = cos (0) - (0)

f(0) = 1 - 0

f(0) = 1 > 0

f(1)  = cos (1) - 1³

f(1) = 0.5403 - 1

f(1) = -0.4597

f(1) = - 0.46

f(1) = - 0.46 < 0

Since, 1 > 0 > -0.46, thus there is a number ''c" in (0,1)

such that f(c) = 0

By applying the Intermediate Value Theorem, there is a root of the equation in the interval (0,1)

b) Given that:

The interval length = 0.01; this implies that it is 0.005 length from its root.

f(0.865) = 0.0014

f(0.87) = - 0.013

The solution to the interval lies between 0.865 , 0.87 by using a calculator at length 0.01.

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right