The intervals in which the function $f(x)=2x^3+9x^2+12x+20$ is increasing is given by :
Answer & explanation
Correct answer: option 3
Given function: $f(x)=2x^{3}+9x^{2}+12x+20$
Differentiate: $f'(x)=\frac{d}{dx}\big(2x^{3}+9x^{2}+12x+20\big)=6x^{2}+18x+12$
Factorize the derivative: $f'(x)=6(x^{2}+3x+2)=6(x+1)(x+2)$
Critical points are roots of $f'(x)$: $x=-2$ and $x=-1$.
Sign analysis of $f'(x)=6(x+1)(x+2)$:
For $x<-2$: $(x+2)<0,\ (x+1)<0 \Rightarrow (x+1)(x+2)>0 \Rightarrow f'(x)>0$
For $x>-1$: $(x+2)>0,\ (x+1)>0 \Rightarrow (x+1)(x+2)>0 \Rightarrow f'(x)>0$
Therefore, $f$ is increasing on $(-\infty,-2)\cup(-1,\infty)$ and decreasing on $(-2,-1)$.
Answer: $(-\infty,-2)\cup(-1,\infty)$