1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The absolute maximum and minimum value of f(x)=3x4−8x3+12x2−48x+25,x∈[0,3] are respectively

A
25,39
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
25,39
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
8,8
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
8,10
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B 25,−39Differentiate the function f(x)=3x4−8x3+12x2−48x+25 with respect to x, f′(x)=12x3−24x2+24x−48 Put f′(x)=0, 12x3−24x2+24x−48=0 (x−2)(12x2+24)=0 (x−2)12(x2+2)=0 x−2=0 x=2 Or, x2=−2 x=√2i But √2i is an imaginary number, so the value to be considered is x=2. Now, substitute the value of x and the end points in the given function. f(0)=3(0)4−8(0)3+12(0)2−48(0)+25 =25 f(2)=3(2)4−8(2)3+12(2)2−48(2)+25 =−39 f(3)=3(3)4−8(3)3+12(3)2−48(3)+25 =10 Therefore, the maximum value is 25 and the minimum value is −39.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Extrema
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program