Thank you for reporting, we will resolve it shortly
Q.
For real x, let $f (x) = x^3 + 5x + 1$, then
Relations and Functions - Part 2
Solution:
Given that $f (x) = x^3 + 5x + 1$
$\therefore \, f ' (x) = 3x^2 + 5 > 0, \forall x \in R$
$\Rightarrow \, f (x)$ is strictly increasing on R
$\Rightarrow \, f (x)$ is one one
$\therefore $ Being a polynomial f (x) is continuous and increasing.
on R with $\displaystyle\lim_{x \to \infty } f(x) = - \infty $ and $\displaystyle\lim_{x \to \infty} f(x) = \infty$
$\therefore $ Range of $f = (- \infty , \infty) = R$
Hence f is onto also. So, f is one one and onto R.