range of $3x^2-2xy$ subjected to $x^2+y^2=1$ If $x^2+y^2=1$. then the range of expression $3x^2-2xy$ without trigonometric substitution method
what i have done  try here is  use arithmetic geometric inequality
$\displaystyle x^2+y^2\geq 2xy$
$\displaystyle -2xy\geq -(x^2+y^2)$
$\displaystyle 3x^2-2xy\geq 2x^2-y^2$
this will not help more
how do i solve it help  me please

The range that you're after is $\left[\frac{3-\sqrt{13}}2,\frac{3+\sqrt{13}}2\right]$. In fact, if $x^2+y^2=1$, then$$3x^2-2xy=3x^2\pm x\sqrt{1-x^2}.$$So, for each $x\in[-1,1]$, let$$f(x)=3x^2-2x\sqrt{1-x^2}\text{ and let }g(x)=3x^2+x\sqrt{1-x^2}.$$Now, using that standard Calculus techniques, you can check that the range of both functions $f$ and $g$ is $\left[\frac{3-\sqrt{13}}2,\frac{3+\sqrt{13}}2\right]$.
