How many real solutions does the equation
\[{x^7} + 14{x^5} + 16{x^3} + 30x - 560 = 0\] have?
Solutions · 1
Official solution
Let \[f\left( x \right) = {x^7} + 14{x^5} + 16{x^3} + 30x - 560\]
\[ \Rightarrow f'\left( x \right) = 7{x^6} + 70{x^4} + 48{x^2} + 30 > 0,\,\forall x \in R\]
\[ \Rightarrow f\] is an increasing function on R
Also \[\mathop {\lim }\limits_{x \to \infty } \,\,f\left( x \right) = \infty \] and \[\mathop {\lim }\limits_{x \to - \infty } \,\,f\left( x \right) = - \infty \]
\[ \Rightarrow \] The curve \[y = f\left( x \right)\] crosses x-axis only once.
\[\therefore\] \[f\left( x \right) = 0\] has exactly one real root.