Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
2.8k views
in Derivatives by (49.2k points)
closed by

The function f(x) = x3+6x2+15x - 12 is

A. strictly decreasing on R

B. strictly increasing on R

C. increasing in (-∞, 2) and decreasing in (2, ∞)

D. none of these

1 Answer

+1 vote
by (50.7k points)
selected by
 
Best answer

Answer is : B. strictly increasing on R

Given:

f(x) = x3+6x2+15x -12.

f’(x) = 3x2+12x+15

f’(x) = 3x2+12x+12+3

f’(x) = 3(x2+4x+4)+3

f’(x) = 3(x+2)2+3

As square is a positive number

∴ f’(x) will be always positive for every real number

Hence f’(x) >0 for all x ϵ R

∴ f(x) is strictly increasing.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...