To check if the function is one-one, we need to analyze its derivative. If the derivative is always positive or always negative, the function is one-one.
The derivative of \(f(x) = x^3 + 5x + 1\) is \(f'(x) = 3x^2 + 5\).
Since \(x^2\) is always non-negative for real x, \(3x^2\) is also non-negative. Therefore, \(3x^2 + 5\) is always greater than or equal to 5, which means \(f'(x) > 0\) for all real x. This implies that the function is strictly increasing.
Since the function is strictly increasing, it is one-one.
To check if the function is onto, we need to see if the range of the function is equal to the codomain (which is R in this case). Since \(f(x)\) is a polynomial of odd degree, its range is R.
As \(x\) approaches \(-\infty\), \(f(x)\) approaches \(-\infty\), and as \(x\) approaches \(\infty\), \(f(x)\) approaches \(\infty\). Also, since \(f(x)\) is continuous, it takes on all real values. Therefore, the function is onto.
The function \(f(x)\) is both one-one and onto on R.
Final Answer: f is one-one and onto on R
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