The function $$f(x) = \int_0^{x^2} \frac{t^{2} -5t + 4}{2+e^{t}} \,dt$$ has two Maxima and Three Minima Points.
I have successfully found the extreme points $x = 0,\pm 1,\pm 2$ but from this how can I conclude that function has two Maxima and three Minima? ( Obviously one way is to take second order derivative and substitute the points to get max / min but this seems pretty lengthy ) Is there any other way of knowing it or we just have to stick with $2$nd derivative test ?
Here by Lebnitz rule we have $f'(x)=2x \frac{(x^2-1)(x^2-4)}{2+e^{x^2}}.$ Next $f'(x)=0$, gives $x=0,\pm 1, \pm 2$.
If $f'(a)=0$: (1) $f'(a-h)>0$ and $f'(a+h)<0$, then these exists a local max at $x=a$. (2) $f'(a-h)<0$ and $f'(a+h)>0$, there exists a local min at $x=a$. With this sign rule, here local max occurs at $x=\pm 1$ and local min occurs at $x=0,\pm 2$.
Edit: When highest/lowest value of f(x) exists at one fixed point in the domain, it is local max/min. If it exists at infinity it becomes a non -local max/min, for example in f(x)=tanhx they exist at infinity. A non local one may exist at infinitely many points. For example g(x)=|x-1|+|x+1|, the lowest value of 2 occurs in the internal [-1,1]. Some time a non local one occurs at a point outside the domain for example h(x)= sim x, x in (-pi/2,pi/2), notice that -pi/2 and pi/2 are not in the domain. Non local max/min are known by their height /depth and not by any position. They are called supremum and infimum.
By global max/min in the interval [a,b] max/min of all local max/min where f(a) and f(b) values also compete.
Global max/min M/m on the interval (a,b) is the max/min of all local max/min, provided M >(f(a),f(b)) or m< (f(a),f(b)).