Question

Cho hàm số \(-\dfrac{1}{3}x^3+2x^2+\left(2m+1\right)x-3m+2.\) Tìm \(m\) để:

a) \(y'\le0,\forall x\in R.\)

b) Tìm \(m\) để \(y'\ge0,\forall x\in R\) với \(y=\dfrac{1}{3}x^3+mx^2-mx+1\).

Answer 1

a: \(y=-\dfrac{1}{3}x^3+2x^2+\left(2m+1\right)x-3m+2\)

=>\(y'=-\dfrac{1}{3}\cdot3x^2+2\cdot2x+\left(2m+1\right)=-x^2+4x+\left(2m+1\right)\)

Để y'<=0 với mọi x thì

\(\left\{{}\begin{matrix}\text{Δ}< =0\\a< 0\end{matrix}\right.\)

=>\(4^2-4\cdot\left(-1\right)\left(2m+1\right)< =0\)

=>16+4(2m+1)<=0

=>4(2m+1)<=-16

=>2m+1<=-4

=>2m<=-5

=>\(m< =-\dfrac{5}{2}\)

b: \(y=\dfrac{1}{3}x^3+mx^2-mx+1\)

=>\(y'=\dfrac{1}{3}\cdot3x^2+m\cdot2x-m=x^2+2m\cdot x-m\)

Để y'>=0 với mọi x thì \(\left\{{}\begin{matrix}\text{Δ}< =0\\a>0\end{matrix}\right.\)

=>\(\left(2m\right)^2-4\cdot1\cdot\left(-m\right)< =0\)

=>\(4m^2+4m< =0\)

=>m(m+1)<=0

=>-1<=m<=0