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Consider the function f(x)=sin(x) in the interval x=[π/4,7π/4]. The number and location(s) of the local minima of this function are

(A) One, at π/2
(B) One, at 3π/2
(C) Two, at π/2 and 3π/2
(D) Two, at π/4 and 3π/2

Please explain how pie/4 is local minima ?  I mean when do we consider extreme points for local minima? in books they consider only for f'(x) =0 that is points where slope is 0 for f(x) not the end points.End points is used for absolute minima/maxima ( as much as i have read in books that is).
closed as a duplicate of: GATE2012-9
asked in Mathematical Logic by (195 points)
closed by | 112 views
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The trick is local minima. Most of us miss this and choose global minima in that range as 3*pi/2 as the answer. But the result should be D as we are talking about local minima.

There are 2 local minima points pi/4 and 3*pi/2

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Yes but what i can't understand is why is pie/4 considered as a "Minima" .I mean with respect to what it is minima?with respect to the f'(x) = 0 at pie/2 (Meaning as there is a maxima at pie/2 and hence the farthest point pie/4 is local minima to it? ) Coz for f'(x) =0 we can check minima or not using f"(x) > 0 but that rule we cannot apply for this extreme points as f'(x) not equal to 0.
0
It's local minima with respect to the f'(x) = 0 at pie/2

1 Answer

+1 vote
answered by Active (2.4k points)
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Thanks!

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