Answer:
Explanation:
To prove that the inequality 2^x ≥ x+1 is false for all positive real numbers x, we can use proof by contradiction.
Assume that the inequality 2^x ≥ x+1 is true for some positive real number x. Then we can write:
2^x ≥ x+1
Taking the logarithm base 2 of both sides of the inequality, we get:
x ≥ log2(x+1)
Now, consider the function f(x) = x - log2(x+1). Taking the derivative of f(x) with respect to x, we get:
f'(x) = 1 - (1 / (ln2 * (x+1)))
Since x > 0, we have x+1 > 1, and therefore (ln2 * (x+1)) > 0. Hence, 1 / (ln2 * (x+1)) > 0, which implies that f'(x) > 0 for all positive real numbers x. Therefore, the function f(x) is strictly increasing for all positive real numbers x.
Now, we know that f(0) = 0 - log2(1) = 0, and f(1) = 1 - log2(2) = 0. Therefore, by the intermediate value theorem, there exists a value c in the interval (0, 1) such that f(c) = 0. But this implies that:
c - log2(c+1) = 0
which contradicts the assumption that 2^x ≥ x+1 for all positive real numbers x. Therefore, the inequality 2^x ≥ x+1 is false for all positive real numbers x.