How to find the maxima? This is a simple question : Find the maximum value of $\frac { 1 }{ { x }^{ 2 }-6x+2 }$
I rewrote ${ x }^{ 2 }-6x+2$ as $(x-3)^{2} - 7$, now when this is min, the original function is max, thus the answer should be $\frac { -1 }{ 7 } $ .
Now a friend asked me that what's wrong in putting $x=0$ in the original equation, you'll get max as $\frac { 1 }{ 2 } $ OR when the denominator is $0$, the max will be infinity and I couldn't give him a proper explanation.
Can someone help me?
 A: There is no maximum value of the function $\dfrac{1}{x^2-6x+2}$, since when $(x-3)^2-7$ is small positive, the reciprocal is huge.  And we can make $(x-3)^2-7$ as small positive as we wish, by taking $x$ a tiny bit larger than $3+\sqrt{7}$ or a tiny bit smaller than $3-\sqrt{7}$.
We do have a local maximum at $x=3$, since as we move away from $x=3$ in either direction for a while, the function $\dfrac{1}{x^2-6x+2}$ decreases. But, as your friend pointed out, we do not have a global maximum at $x=3$.
Remark: In order to see what's happening, nothing is as good as a picture. Use Wolfram Alpha, or some other graphing program, or a graphing calculator, to draw a picture of the curve $y=\dfrac{1}{(x-3)^2-7}$. Everything will become clear.
A: Let $$y=\frac1{x^2-6x+2}$$
$$\implies y\cdot x^2-6y\cdot x+2y-1=0$$
As $x$ is real, the discriminant of the above Quadratic equation must be $\ge0$
i..e, $$(-6y)^2-4y(2y-1)\ge0\implies y(7y+1)\ge0$$
Either $y\ge0$ and $y\ge -\frac17\implies y\ge0$
or  $y\le0$ and $y\le -\frac17\implies y\le -\frac17$
$$\implies -\infty <y\le -\frac17\cup 0\le y<\infty$$
