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I know how local & absolute maxima or minima are calculated, but I don't understand that if we differentiate a function once & put it equal to zero, we get $1,2$ or any other limited number of points of maxima or minima.

But as we all know that there are many local maxima/minima & at each one of them, the slope of function changes its sign(from + to - or converse), so it's value become zero at them. Then why is it that the first derivative equation gives us only few of them? I mean if slope is zero at all of them, then they must all satisfy the equation (I know that the number of roots given by first derivative equation would depend on it being quadratic or cubic or of whatever degree it is, and this makes it even more ambiguous to me).

I know it may sound absurd to some, but I am just a beginner at calculus and have searched quite many places on the web but found nothing. Please help!

hardmath
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shashank tyagi
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  • There are functions that are not differentiable, and then the derivative won't be zero. The standard example is $|x|$, which clearly has a global minimum. – Arthur Jul 25 '16 at 16:57
  • It is not clear what you are asking. If a function is differentiable at a local maximum, then the derivative will be zero there. So, if you are looking for local maxima, you can look for all places where the derivative is zero and then see which are actual local maxima. – copper.hat Jul 25 '16 at 17:30
  • Yes I know but I am talking about functions which are differentiable. @Arthur – shashank tyagi Jul 25 '16 at 17:30
  • Ok, I get it for sinx. But let's consider another function(Its a question I saw in a book):- Find the maxima value of (f(x)= x³-9x²+24x+5) in the interval [1,6]. Now in this question, the local maxima is equal to 2(at which the value of function is 25), But as we find absolute maxima by method:- Absolute Maxima= Max{ f(1), f(6), all its local maxima in [1,6] }. We find that 6 gives the absolute maximum value, which means 6 is also a local maxima(as absolute maxima is also a local maxima among many local maxima). Then why isn't 6 a solution of the function's first derivative? – shashank tyagi Jul 25 '16 at 17:37

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We can get as many (local) maxima and minima as we want - the number is not limited. Consider $\sin x$ on $\mathbb{R}$. It has infinitely many maxima and minima (and at all of them the derivative, $\cos x$, is zero).

smcc
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  • Ok, I get it for sinx. But let's consider another function(Its a question I saw in a book):- Find the maxima value of (f(x)= x³-9x²+24x+5) in the interval [1,6]. Now in this question, the local maxima is equal to 2(at which the value of function is 25), But as we find absolute maxima by method:- Absolute Maxima= Max{ f(1), f(6), all its local maxima in [1,6] }. We find that 6 gives the absolute maximum value, which means 6 is also a local maxima(as absolute maxima is also a local maxima among many local maxima). Then why isn't 6 a solution of the function's first derivative? – shashank tyagi Jul 25 '16 at 17:25
  • You can only use the first derivative test on open sets, if you have a constraint of the form $1 \le x \le 6$ then the maximum (or minimum) may occur at $x=1$ or $x=6$ but the derivative need not be zero there. For example, with the function $x \mapsto x$, on the interval $[1,6]$, will have a maximum value at $x=6$, but clearly the first derivative is non zero there. – copper.hat Jul 25 '16 at 17:37