Question

Find both the maximum value and the minimum value of 3 x 4 − 8 x 3 + 12 x 2 − 48 x + 25 on the interval [0, 3]

Answer 1

Given function is,

f( x )=3 x 4 −8 x 3 +12 x 2 −48x+25

Differentiate the function with respect to x,

f ′ ( x )=12 x 3 −24 x 2 +24x−48

Put f ′ ( x )=0,

12 x 3 −24 x 2 +24x−48=0 12 x 2 ( x−2 )+24( x−2 )=0 ( 12 x 2 +24 )( x−2 )=0 x=2,± −2

As x= −2 is not a real number, so, x=2 is the only critical point of the given function.

Since the interval is given as [ 0,3 ], so, the value of f( x ) is calculated at the critical point and at the end points of the interval that is, 0, 2 and 3.

At x=0,

f( 0 )=3 ( 0 ) 4 −8 ( 0 ) 3 +12 ( 0 ) 2 −48( 0 )+25 =25

At x=2,

f( 2 )=3 ( 2 ) 4 −8 ( 2 ) 3 +12 ( 2 ) 2 −48( 2 )+25 =−39

At x=3,

f( 3 )=3 ( 3 ) 4 −8 ( 2 ) 3 +12 ( 3 ) 2 −48( 3 )+25 =16

Therefore, minimum value of the given function is −39 which is at x=2 and maximum value of the function is 25 which is at x=0.