What is the maximum of $\frac{1}{1+x^2+2x}$. I think it should be $1$ since we can set $x=0$ and if we make $x$ any other value it will increase the denominator which messes everything up and makes it less.
But my teacher told me I was wrong and no global max exists How? I know the min is zero as $0$ approaches infinity.
On
Minimize the value of the denominator:
$1+2x+x^2=(x+1)^2$. This means that at $x=1$ your quadratic is the smallest but if you plug in $x=-1$ you will get zero and you can’t divide by zero so the function will have an asymptote (vertical) and approach infinity from both sides.
This means that as you approach $-1$, your denominator goes to zero so your function gets bigger and bigger so your function does not have a global maximum and the closer we get to $x=-1$ the bigger it gets (but it is not defined at that point).
It has no maximum, since $$\lim _{x \to -1} f(x) =\lim _{x \to -1} {1\over (x+1)^2} = \infty$$