Question

Find the greatest value of $x$ in order that $7x^2+11$ may be greater than $x^3+17x$.

Answer 1

$7x^2+11$ $<$ $x^3+17x$

$=>x^3-7x^2+17x-11$ $<$ $0$

$=>x^3-6x^2+11x-x^2+6x-11$ $<$ $0$

$=>x(x^2-6x+11)-1(x^2-6x+11)$ $<$ $0$

$=>(x-1)(x^2-6x+11)$ $<$ $0$

$=>(x-1)\left[\right(x-3{)}^2+2]$$<$ $0$

we know $(x-3{)}^2+2>0$ always,so the condition is

$x-1$$<$ $0$

$x$$<$ $1$

thus $x=0$ if $x$ is an integer.