Chapter 5: Problem 28
Prove that the equation \(3 x^{5}+15 x-8=0\) has only one real solution.
Short Answer
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The derivative of the function \(f(x) = 3x^5 + 15x - 8\) is \(f'(x) = 15x^4 + 15\), which is always positive for all real values of x. Since \(f'(x) > 0\), the function is monotonically increasing on the entire real number domain. A monotonically increasing function can have only one zero crossing on the domain. Therefore, the equation \(3x^5 + 15x - 8 = 0\) has only one real solution.
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