Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Applications of Derivatives

Question:

In which interval the function $f(x)=\frac{1}{3}x^3+\frac{1}{2}x^2-6x+8$ decreases?

Options:

$x < – 2$

$x > 2$

$– 3 < x < 2$

No where

Correct Answer:

$– 3 < x < 2$

Explanation:

$f ' x = x + x − 6 <0$

f(x) is a decreasing function if $f ' x <0$

$⇒x^2+x-6<0$

$⇒(x+3)(x-2)<0$

$– 3 < x < 2$