Use the intermediate value theorem to prove that the equation x^3=x+8 has at least one solution

Question

Use the intermediate value theorem to prove that the equation x^3=x+8 has at least one solution

Answer 1

The function x^3=x+8 can be written as x^3-x-8=0

For f(3): (27 -(3) -8) = 16

 

For f(2): (8 -(2) -8) = -2

 

For f(0): -8

For f(-2): (-8 -(-2) -8) = -14

Since f(3) = 16, and f(2) = -2 and f(-2) = -14, there will be some value x such that f(x) = 0. And, since the function is a polynomial, it will be continuous in the interval [-2,3]. Therefore, the function x^3=x+8 does, in fact, have a real solution.

 

Using the Bisection method we converge on a solution by iteratively bisecting (cutting in half) an upper and lower value starting with f(-2) and f(3). Doing so, our solution is x = 2.166312754. An advanced graphing calculator such as the TI-83, 84 or 89 would be an asset in solving such problems.

Answer 2

First take all terms to one side, x³-x-8=0.  Find f(x) by setting the it equal to the left expression, f(x) = x³-x-8.  Find one x-value where f(x) < 0 and a second x-value where f(x)>0 by inspection or a graph.  f(2) = -2 and f(3) = 16.  Since f(x) is a continuous function (all polynomials are continuous) on [2,3] and 0 is between -2 and 16 then there exists a number c in (2,3) such that f(c) = 0.

 

A less formal way of stating the last sentence is "Since f(x) is a continuous function on [2,3] and f(2) and f(3) have opposite signs the graph of f(x) must cross the x-axis."

Answer 3

f(1) = 1^3 - 1 - 8

f(1) = -8

 

f(3) = 3³ - 3 - 8

f(3) = 16

 

∃ w | 1 < w < 3 and f(w) = 0

Answer 4

Consider the function f(x) = x³ -x - 8

This function is continuous

 

Let x = 2. Then f(2) = -1

Let x = 3. Then f(3) = 16

 

By the intermediate value theorem, on the interval

[2, 3], for f(2) < c < f(3) there is at least one x

such that f(x) = c.

Using the evaluated function values c would be in the

range -1 < c < 16.

 

Since 0 is in this interval, when c = 0, then f(x) = 0

has at least one solution, then x ³ = x + 8 has at

least one solution.