GATE CSE
First time here? Checkout the FAQ!
x
+2 votes
161 views

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 (3.8k points)  
retagged by | 161 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 (963 points)  
the value at 0 is one right?
@Tehreem at x = 0
f(x) = 2


Top Users Aug 2017
  1. Bikram

    4902 Points

  2. ABKUNDAN

    4704 Points

  3. akash.dinkar12

    3480 Points

  4. rahul sharma 5

    3158 Points

  5. manu00x

    3012 Points

  6. makhdoom ghaya

    2480 Points

  7. just_bhavana

    2388 Points

  8. stblue

    2138 Points

  9. Tesla!

    2060 Points

  10. joshi_nitish

    1758 Points


25,014 questions
32,140 answers
74,824 comments
30,185 users