MTH101 Solved Practice Questions Lectures 23-45 Set 2

MTH101 Practice Questions/ Solutions · Lecture No. 23 to 45 · Lecture No. 23: Maximum and Minimum Values of Functions

Q1. Find the minimum value of the function

f x x x attains in the interval [4,-4].

Solution:

First of all wehavetofindcriticalpointsbyputting '( )=0

3( 9) 3,3

(4) (4) 27(4 f x f x x x f x x put f x x x x x

f ( 2) ( 4) 27( 4) (3) (3) 27(3) ( 3) ( 3) 27( 3) Sominimum value f f f

Q2. Find the maximum value of the function

f x x x x attains in the interval [-4, 3].

Solution:

First of all wehavetofindcriticalpointsbyputting

3) 3)

1. ( 3) 1, f x f x

x x x f x x x put f x x x x x x x x x x x x x x x (1) (1) 3(1) 9(1) ( 3) ( 3) 3( 3) 9( 3) (3) (3) 3(3) 9(3) ( 4) ( 4) 3( 4) 9( 4) Hencemaximum value ( 3) (3) f f f f f f

Q3. Find the absolute maximum and absolute minimum values of the function f x

x on interval 3 x

Solution:

First wewillfindcriticalpoints: put

Now wefind valueof ( )atcriticalpointandat theend pointsof interval : (0) (0) (1) (1) ( 3) ( 3) Henceabsolutemaximum=4andabsoluteminimum=-5 f x x x f x f f f

Q4. Find the absolute maximum and absolute minimum values of the function f x

x on interval 2 x

Solution:

First wewillfindcriticalpoints: (x) 1,sothereisnocriticalpoints. Now wefind valueof ( ) at theend pointsof interval : ( 2) (2) Henceabsolutemaximum=4andabsoluteminimum=0 Since f f x f f

Q5. Find the maximum and minimum value of the function f x

x x + on the interval (

Solution:

This is a continuous function on the given interval and lim lim (3 1)

lim lim (3 1) x x

x x f x x x

f x x x So f has a minimum but no maximum value in the interval ( −∞+∞ . To find the minimum value put f x = i.e 12 ( 4) 2 are thecriticalpoints. At = 0 , (0) 1, at 2, (2) 3(2) -24(2) +1=-47 and at -2 , ( 2) 3( 2) -24(-2) +1=-47 so minimum value occur f x x x x x x and x x f x f x f s at = 2 and it is equal to ' ( ) -47' x f x