Is there a number that is exactly 5 more than its cube?
When we multiply a number three times by itself, the resultant number is known as the cube of the original number.
Answer: Yes, there is a number that is exactly 5 more than its cube.
We will prove that there exists a number that is exactly 5 more than its cube by using the Intermediate Value Theorem. The IVT states that if a function f is continuous on an interval [a,b] and if f(a) < 0 and f(b) > 0 (or vice-versa), then there is a third point c with a < c < b so that f(c) = 0.
Let x be the number.
x = x3 + 5
Let function f(x) = x3 - x + 5
Since f(-2) = -8 + 2 + 5 = -1 < 0, f(0) = 5 > 0, and the function f(x) is continuous in the interval [-2, 0], using intermediate value theorem (IVT) there is at least one point c in (-2, 0) such that f(c) = 0.
⇒ c3 - c + 5 = 0
⇒ c = c3 + 5