In which interval the function $f(x)=\frac{1}{3}x^3+\frac{1}{2}x^2-6x+8$ decreases?
Answer & explanation
Correct answer: option 3
$f ' x = x + x − 6 <0$
f(x) is a decreasing function if $f ' x <0$
$⇒x^2+x-6<0$
$⇒(x+3)(x-2)<0$
$– 3 < x < 2$
In which interval the function $f(x)=\frac{1}{3}x^3+\frac{1}{2}x^2-6x+8$ decreases?
Correct answer: option 3
$f ' x = x + x − 6 <0$
f(x) is a decreasing function if $f ' x <0$
$⇒x^2+x-6<0$
$⇒(x+3)(x-2)<0$
$– 3 < x < 2$