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

The interval of f(x)=x3+2x2+5x,x<0 for which it is concave upwards is

A
(0,)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(23,0)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(23,)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(,0)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B (23,0)
f(x)=x3+2x2+5x, x<0 (i)
f(x)=3x2+4x+5
f′′(x)=6x+4
For f to be concave upward,
f′′(x)>0
x>23 (ii)
From (i) and (ii), we get
Interval of concave upwards is (23,0)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon