For real x, let f(x) = x^3 + 5x + 1, then | vedclass

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(C) Given $f(x) = x^3 + 5x + 1.$First,we find the derivative: $f'(x) = 3x^2 + 5.$Since $x^2 \ge 0$ for all $x \in R,$ it follows that $3x^2 + 5 \ge 5

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$f$ is onto and one-one $R$

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(C) Given $f(x) = x^3 + 5x + 1.$First,we find the derivative: $f'(x) = 3x^2 + 5.$Since $x^2 \ge 0$ for all $x \in R,$ it follows that $3x^2 + 5 \ge 5 > 0$ for all $x \in R.$Because $f'(x) > 0$ for all $x,$ the function $f(x)$ is strictly increasing on $R.$$A$ strictly increasing function is always one-one.Next,we check if the function is onto. Since $f(x)$ is a polynomial of odd degree,its range is $(-\infty, \infty).$Specifically,$\lim_{x 	o -\infty} f(x) = -\infty$ and $\lim_{x 	o \infty} f(x) = \infty.$Since $f(x)$ is continuous and its range is the set of all real numbers $R,$ the function is onto.Therefore,$f$ is both one-one and onto.

For real $x,$ let $f(x) = x^3 + 5x + 1,$ then

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