wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The least value of the function f(x) = x3-18x2+96x in the interval [0,9] is
(a) 126
(b) 135
(c) 160
(d) 0

Open in App
Solution

(d)0Given:fx=x3-18x2+96x⇒f'x=3x2-36x+96Foralocalmaximaoralocalminima,wemusthavef'x=0⇒3x2-36x+96=0⇒x2-12x+32=0⇒x-4x-8=0⇒x=4,8So,f8=83-1882+968=512-1152+768=128f4=43-1842+964=64-288+384=160f0=03-1802+960=0f9=93-1892+969=729-1458+864=135Hence, 0 is the minimum value in the range0,9.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Organisation of Data
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon
footer-image