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\(y'=x^2-2\left(m-1\right)x+3\left(m-1\right)\)
Hàm đồng biến trên khoảng đã cho khi với mọi \(x>1\) ta luôn có:
\(g\left(x\right)=x^2-2\left(m-1\right)x+3\left(m-1\right)\ge0\)
\(\Rightarrow\min\limits_{x>1}g\left(x\right)\ge0\)
Do \(a=1>0;-\dfrac{b}{2a}=m-1\)
TH1: \(m-1\ge1\Rightarrow m\ge2\)
\(\Rightarrow g\left(x\right)_{min}=f\left(m-1\right)=\left(m-1\right)^2-2\left(m-1\right)^2+3\left(m-1\right)\ge0\)
\(\Rightarrow\left(m-1\right)\left(4-m\right)\ge0\Rightarrow1\le m\le4\Rightarrow2\le m\le4\)
TH2: \(m-1< 1\Rightarrow m< 2\Rightarrow g\left(x\right)_{min}=g\left(1\right)=m\ge0\)
Vậy \(0\le m\le4\)
Hàm số xác định trên R khi và chỉ khi:
\(sin^2x+\left(2m-3\right)cosx+3m-2>0;\forall x\in R\)
\(\Leftrightarrow-cos^2x+\left(2m-3\right)cosx+3m-1>0\)
\(\Leftrightarrow t^2-\left(2m-3\right)t-3m+1< 0;\forall t\in\left[-1;1\right]\)
\(\Leftrightarrow t^2+3t+1< m\left(2t+3\right)\)
\(\Leftrightarrow\dfrac{t^2+3t+1}{2t+3}< m\) (do \(2t+3>0;\forall t\in\left[-1;1\right]\))
\(\Leftrightarrow m>\max\limits_{\left[-1;1\right]}\dfrac{t^2+3t+1}{2t+3}\)
Ta có: \(\dfrac{t^2+3t+1}{2t+3}=\dfrac{t^2+t-2+2t+3}{2t+3}=\dfrac{\left(t-1\right)\left(t+2\right)}{2t+3}+1\)
Do \(-1\le t\le1\Rightarrow\dfrac{\left(t-1\right)\left(t+2\right)}{2t+3}\le0\)
\(\Rightarrow\max\limits_{\left[-1;1\right]}\dfrac{t^2+3t+1}{2t+3}=1\)
\(\Rightarrow m>1\)
\(y'=-\dfrac{2x-2}{\left(x^2-2x+5\right)^2}=\dfrac{2-2x}{\left(x^2-2x+5\right)^2}\)
\(y'\ge0\Leftrightarrow\dfrac{2-2x}{\left(x^2-2x+5\right)^2}\ge0\Rightarrow x\le1\)
Có \(1-\left(-8\right)+1=10\) số nguyên
cho hàm số \(y=\dfrac{x^2+mx-3}{x+2}\) (m la tham số). biết \(y'\left(-1\right)=4\). tính giá trị m?
\(y'=\dfrac{\left(2x+m\right)\left(x+2\right)-\left(x^2+mx-3\right)}{\left(x+2\right)^2}=\dfrac{x^2+4x+2m+3}{\left(x+2\right)^2}\)
\(y'\left(-1\right)=\dfrac{2m}{1}=2m=4\Rightarrow m=2\)
1. \(y'=3x^2\sqrt{x}+\dfrac{x^3-5}{2\sqrt{x}}=\dfrac{7x^3-5}{2\sqrt{x}}\)
2. \(y'=3x^5+\dfrac{3}{x^2}+\dfrac{1}{\sqrt{x}}\)
3. \(y'=2-\dfrac{2}{\left(x-2\right)^2}\)
y = \(\dfrac{sin^2x}{cosx\left(sinx-cosx\right)}+\dfrac{1}{4}\)
y = \(\dfrac{sin^2x}{sinx.cosx-cos^2x}+\dfrac{1}{4}=\dfrac{\dfrac{sin^2x}{cos^2x}}{\dfrac{sinx.cosx}{cos^2x}-1}+\dfrac{1}{4}\)
y = \(\dfrac{tan^2x}{tanx-1}+\dfrac{1}{4}\)
y = \(\dfrac{4tan^2x+tanx-1}{4tanx-4}\). Đặt t = tanx. Do x ∈ \(\left(\dfrac{\pi}{4};\dfrac{\pi}{2}\right)\) nên t ∈ (1 ; +\(\infty\))\
Ta đươc hàm số f(t) = \(\dfrac{4t^2+t-1}{4t-4}\)
⇒ ymin = \(\dfrac{17}{4}\) khi t = 2. hay x = arctan(2) + kπ
a: \(y=-x^3-\left(m+1\right)x^2+3\left(m+1\right)x\)
=>\(y'=-3x^2-\left(m+1\right)\cdot2x+3\left(m+1\right)\)
=>\(y'=-3x^2+x\cdot\left(-2m-2\right)+\left(3m+3\right)\)
Để hàm số nghịch biến trên R thì \(y'< =0\forall x\)
=>\(\left\{{}\begin{matrix}\text{Δ}< =0\\a< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(-2m-2\right)^2-4\cdot\left(-3\right)\left(3m+3\right)< =0\\-3< 0\end{matrix}\right.\)
=>\(4m^2+8m+4+12\left(3m+3\right)< =0\)
=>\(4m^2+8m+4+36m+36< =0\)
=>\(4m^2+44m+40< =0\)
=>\(m^2+11m+10< =0\)
=>\(\left(m+1\right)\left(m+10\right)< =0\)
TH1: \(\left\{{}\begin{matrix}m+1>=0\\m+10< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>=-1\\m< =-10\end{matrix}\right.\)
=>\(m\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}m+1< =0\\m+10>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< =-1\\m>=-10\end{matrix}\right.\)
=>-10<=m<=-1
b: \(y=-\dfrac{1}{3}x^3+mx^2-\left(2m+3\right)x\)
=>\(y'=-\dfrac{1}{3}\cdot3x^2+m\cdot2x-\left(2m+3\right)\)
=>\(y'=-x^2+2m\cdot x-\left(2m+3\right)\)
Để hàm số nghịch biến trên R thì \(y'< =0\forall x\)
=>\(\left\{{}\begin{matrix}\text{Δ}< =0\\a< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-1< 0\\\left(2m\right)^2-4\cdot\left(-1\right)\cdot\left(-2m-3\right)< =0\end{matrix}\right.\)
=>\(4m^2+4\left(-2m-3\right)< =0\)
=>\(m^2-2m-3< =0\)
=>(m-3)(m+1)<=0
TH1: \(\left\{{}\begin{matrix}m-3>=0\\m+1< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m>=3\\m< =-1\end{matrix}\right.\)
=>\(m\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}m-3< =0\\m+1>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< =3\\m>=-1\end{matrix}\right.\)
=>-1<=m<=3
Hình như là đề sai, hàm số ko có tham số m nào