Determine if there exists a real number that is exactly one less than its cube. If so, find the value of the number. (Hint: IVT)
To determine if there exists a real number that is exactly one less than its cube, we can make use of the Intermediate Value Theorem (IVT).
The IVT states that if a function is continuous on a closed interval [a, b], and takes on values of f(a) and f(b) with opposite signs, then there exists at least one value c in the interval [a, b] such that f(c) = 0.
Let's apply this to our problem. We want to find a real number x such that it is one less than its cube, which can be represented as x = x^3 - 1. Rearranging the equation gives us x^3 - x - 1 = 0.
Now, we can define a function f(x) = x^3 - x - 1. Our goal is to find out if f(x) = 0 has a solution.
Considering the function f(x) = x^3 - x - 1 is a polynomial, it is continuous for all real values of x. Therefore, to apply the IVT, we need to find two points x1 and x2 such that f(x1) and f(x2) have opposite signs.
We can start by substituting a few values into f(x) to determine if there are any changes in sign. Let's evaluate f(0) and f(1):
f(0) = (0^3) - 0 - 1 = -1
f(1) = (1^3) - 1 - 1 = -1
Since both f(0) and f(1) are negative, we know that they have the same sign. Therefore, there is no change in sign between these two points. For simplicity, let's evaluate f(2):
f(2) = (2^3) - 2 - 1 = 5
Now, we notice that f(2) is positive and different in sign from f(0) and f(1). This means that f(x) = 0 must have a root within the interval [0, 2]. Therefore, there exists a real number that is exactly one less than its cube.
To find the exact value of this number, we can solve the equation x^3 - x - 1 = 0. There are various methods to solve this equation, such as numerical methods or using calculus techniques.
One approach is to approximate the root using numerical methods like the Newton-Raphson method or the bisection method. Alternatively, you can use software, online calculators, or computer algebra systems to find the exact value of the root.
In this case, the exact value of the number can be approximated as x ≈ 1.324717957244746.
Therefore, the real number that is exactly one less than its cube is approximately 1.324717957244746.