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Which of the following is the right Procedure to get the minimum for f(x)?

Procedure 1: This is a closed interval, so we will have to calculate the value including and between [0,π/2].

To get critical points we do f'(x)=0. But here on f'(x) we get: -e-x-sin(x)=0, ie, there are no critical point and derivative exists everywhere? So we will calculate the points where derivatives will be 0. Exponential will never be zero, so consider sin, it will be zero only at point nπ/2.(here n=1)

Hence answer is π/2.

Procedure 2: If f'(x)<0, then it is minimum at that point. Here we are getting f'(x)<0 for [0,π/2] and inbetween points. So are we supposed to substitute each value in f(x) from options to check which gives the minimum?

Which Procedure is right?

asked in Calculus by Loyal (3.7k points)  
edited by | 139 views

1 Answer

0 votes
To find maxima or minima we need to equation differential to zero but  the differential is never satisfied by any value in [0,pi/2] so we only need to check if slope is positive or negative to find the maxima , it is  always negative in [0,pi/2] so the value at pi/2 is the minimum.
answered by Junior (943 points)  
the value at 0 is one right?
@Tehreem at x = 0
f(x) = 2


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