Ta có: f′(x)=x2+2(m−1)x−(2m−10)f′(x)=x2+2(m−1)x−(2m−10)
f′(x)>0,∀x∈R⇔x2+2(m−1)x−(2m−10)>0,∀x∈Rf′(x)>0,∀x∈R⇔x2+2(m−1)x−(2m−10)>0,∀x∈R
\(\Leftrightarrow\left\{{}\begin{matrix}a=1>0\\\Delta'=\left(m-1\right)^2+\left(2m-10\right)< 0\end{matrix}\right.\Leftrightarrow m^2-9< 0\Leftrightarrow-3< m< 3\)
Vậy m∈(−3;3)m∈(−3;3) thì f′(x)>0,∀x∈Rf′(x)>0,∀x∈R.
Ta có: f′(x)=x2+2(m−1)x−(2m−10)f′(x)=x2+2(m−1)x−(2m−10)
f′(x)>0,∀x∈R⇔x2+2(m−1)x−(2m−10)>0,∀x∈Rf′(x)>0,∀x∈R⇔x2+2(m−1)x−(2m−10)>0,∀x∈R
⇔\(\left\{{}\begin{matrix}a=1>0\\\Delta'=\left(m-1\right)^2+\left(2m-10\right)< 0\end{matrix}\right.\)⇔m2−9<0⇔m2−9<0 ⇔−3<m<3⇔−3<m<3.
Vậy m∈(−3;3)m∈(−3;3) thì f′(x)>0,∀x∈Rf′(x)>0,∀x∈R.
Ta có: \(f'\left(x\right)=x^2+2\left(m-1\right)x-\left(2m-10\right)\)
\(f'\left(x\right)>0\Leftrightarrow x^2+2\left(m-1\right)-\left(2m-10\right)>0\)
Để f'(x)>0 \(\Leftrightarrow\left\{{}\begin{matrix}a=1>0\\\Delta'=\left(m-1\right)^2+\left(2m-10\right)>0\end{matrix}\right.\)
\(\Leftrightarrow m^2-9< 0\Leftrightarrow-3< m< 3\)
Vậy m∈(-3;3) thì f'(x)>0
Ta có: f′(x)=x2+2(m−1)x−(2m−10)f′(x)=x2+2(m−1)x−(2m−10)
f′(x)>0,∀x∈R⇔x2+2(m−1)x−(2m−10)>0,∀x∈Rf′(x)>0,∀x∈R⇔x2+2(m−1)x−(2m−10)>0,∀x∈R
⇔{a=1>0Δ′=(m−1)2+(2m−10)<0⇔{a=1>0Δ′=(m−1)2+(2m−10)<0
⇔m2−9<0⇔m2−9<0 ⇔−3<m<3⇔−3<m<3.
Vậy m∈(−3;3)m∈(−3;3) thì f′(x)>0,∀x∈Rf′(x)>0,∀x∈R.
f′(x)=x2+2(m−1)x−(2m−10)f′(x)=x2+2(m−1)x−(2m−10)
f′(x)>0,∀x∈R⇔x2+2(m−1)x−(2m−10)>0,∀x∈Rf′(x)>0,∀x∈R⇔x2+2(m−1)x−(2m−10)>0,∀x∈R
⇔{a=1>0Δ′=(m−1)2+(2m−10)<0⇔{a=1>0Δ′=(m−1)2+(2m−10)<0
⇔m2−9<0⇔m2−9<0 ⇔−3<m<3⇔−3<m<3.
f'(x)=x2+2(m-1)x-(2m-10)
f'(x)>0
⇔\(\left\{{}\begin{matrix}a>0\\\Delta'< 0\end{matrix}\right.\)⇔(m-1)2+2m-10<0⇔-3<m<3.
Ta có: f′(x)=x2+2(m−1)x−(2m−10)
f′(x)>0 ⇔x2+2(m−1)x−(2m−10)>0
⇔{a=1>0Δ′=(m−1)2+(2m−10)<0
⇔m2−9<0 ⇔−3<m<3.
Vậy m∈(−3;3) thì f′(x)>0,∀x∈R.
Ta có: f′(x)=x2+2(m−1)x−(2m−10)f′(x)=x2+2(m−1)x−(2m−10)
f′(x)>0,∀x∈R⇔x2+2(m−1)x−(2m−10)>0,∀x∈Rf′(x)>0,∀x∈R⇔x2+2(m−1)x−(2m−10)>0,∀x∈R
⇔{a=1>0Δ′=(m−1)2+(2m−10)<0⇔{a=1>0Δ′=(m−1)2+(2m−10)<0
⇔m2−9<0⇔m2−9<0 ⇔−3<m<3⇔−3<m<3.
Vậy m∈(−3;3)m∈(−3;3) thì f′(x)>0,∀x∈Rf′(x)>0,∀x∈R.
