The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
+2 votes

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 Active (2.9k points)
retagged by | 230 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 (959 points)
the value at 0 is one right?
@Tehreem at x = 0
f(x) = 2

Related questions

0 votes
1 answer
asked Nov 28, 2018 in Calculus by aditi19 Active (3.7k points) | 146 views
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
49,576 questions
54,192 answers
71,147 users