Solve for how many values of $x$, $f(x)=$ $\frac{1000^x}{x!}$ attains the maximum?
I was solving this past problem of an UG entrance. So I found that its go'in to be maximum at $1$ value of n because at one point only its go'in to be undefined.So the maxima exists only at that one value of $x$.But the ans turned out to be $2$ values for $x$.I didn't get it why??
Define $$T_x=\frac{1000^x}{x!}$$ now observe that $$T_{n+1}=\frac{1000}{n+1} \cdot T_n$$ We can notice that the expression is increasing for $0<n \leq 998$ since we have $$\frac{1000}{n+1} \geq 1$$At $n=999$ the multiplier becomes 1 thus $$f(999)=f(1000)$$ Thus there are two points of maxima ..... you can continue this argument for continuous $f(x)$ instead of the discrete $T_n$ I defined...