The differentiable function f (x) defined on R is known, which satisfies f '(x)

If f (x + 1) is an even function and f (2) = 1, f (2) = f (1 + 1) = f (- 1 + 1) = f (0) = 1
When x = 0 is brought into the inequality, we can see that e ^ 0 = 1 = f (0), the inequality is not tenable, so 0 is not the solution of the inequality. When x = 2 is brought into the inequality, we can see that e ^ 2 = 7.389 > F (2) = 1, the inequality is tenable, so 2 is the solution of the inequality, e ^ 4 > 2, so the answer is (0, + ∞)

RELATED INFORMATIONS

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