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\(y=f\left(x\right)=x^4-2x^2\)
\(\Rightarrow f'\left(x\right)=4x^3-4x\)
\(f'\left(x\right)=0\Leftrightarrow4x^3-4x=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=0\end{matrix}\right.\)
\(f\left(1\right)=-1;f\left(-2\right)=8;f\left(-1\right)=-1;f\left(0\right)=0\)
\(\Rightarrow y_{min}=-1;"="\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(y_{max}=8;"="\Leftrightarrow x=-2\)
Đặt \(x^2=t\left(0\le t\le4\right)\)
\(y=f\left(t\right)=t^2-2t\)
\(minf\left(t\right)=min\left\{f\left(0\right);f\left(4\right);f\left(1\right)\right\}=f\left(1\right)=-1\)
\(maxf\left(t\right)=max\left\{f\left(0\right);f\left(4\right);f\left(1\right)\right\}=f\left(4\right)=8\)
\(min=-1\Leftrightarrow x=\pm1\)
\(max=8\Leftrightarrow x=-2\)
Ta có: \(y=\sqrt{3+x}+\sqrt{5-x}\)
ĐKXĐ: \(-3\le x\le5\)
\(y^2=3+x+5-x+2\sqrt{\left(3+x\right)\left(5-x\right)}=8+2\sqrt{\left(3+x\right)\left(5-x\right)}\)\(\ge8\)
\(\Rightarrow y\ge2\sqrt{2}\)
Dấu "=" xảy ra khi và chỉ khi \(\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)(thỏa mãn)
Vậy min y = \(2\sqrt{2}\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
mặt khác \(y^2\) = \(8+2\sqrt{\left(3+x\right)\left(5-x\right)}\le8+3+x+5-x=16\)
\(\Rightarrow y\le4\)
Dấu"=" xảy ra khi và chỉ khi \(3+x=5-x\Leftrightarrow x=1\)(thỏa mãn)
Vậy max y = 4 \(\Leftrightarrow x=1\)
\(h\left(x\right)=x^2-4x+5+m\)
\(g\left(x\right)=\left|h\left(x\right)\right|=\left|f\left(x\right)+m\right|=\left|x^2-4x+5+m\right|\)
\(h\left(0\right)=5+m;h\left(4\right)=5+m;h\left(2\right)=1+m\)
TH1: \(1+m>0\Leftrightarrow m>-1\)
\(max=5+m=9\Leftrightarrow m=4\left(tm\right)\)
TH2: \(5+m< 0\Leftrightarrow m< -5\)
\(max=-1-m=9\Leftrightarrow m=-10\left(tm\right)\)
TH3: \(5+m>0>1+m\Leftrightarrow-5< m< -1\)
Nếu \(5+m< -1-m\Leftrightarrow m< -3\)
\(max=-1-m=9\Leftrightarrow m=-10\left(tm\right)\)
Nếu \(5+m=-1-m\Leftrightarrow m=-3\)
\(max=5+m=2\ne9\)
\(\Rightarrow m=-3\) không thỏa mãn yêu cầu bài toán
Nếu \(5+m>-1-m\Leftrightarrow m>-3\)
\(max=5+m=9\Leftrightarrow m=4\left(tm\right)\)
Vậy \(m=4;m=-10\)
\(y=\left(x+2\right)\left(3-x\right)\)
\(=3x-x^2+6-2x\)
\(=-x^2+x+6\)
=>y'=-2x+1
Đặt y'=0
=>-2x+1=0
=>-2x=-1
=>\(x=\dfrac{1}{2}\)
\(f\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}+2\right)\left(3-\dfrac{1}{2}\right)=\dfrac{5}{2}\cdot\dfrac{5}{2}=\dfrac{25}{4}\)
\(f\left(-2\right)=\left(-2+2\right)\left(3+2\right)=0\)
\(f\left(3\right)=\left(3+2\right)\left(3-3\right)=0\)
=>\(y_{max\left[-2;3\right]}=\dfrac{25}{4}\)
\(f\left(x\right)=x^4-4x^3+4x^2-5x^2+10x-3\)
\(=\left(x^2-2x\right)^2-5\left(x^2-2x\right)-3\)
Đặt \(t=x^2-2x\Rightarrow t\in\left[-1;8\right]\)
Xét hàm \(f\left(t\right)=t^2-5t-3\) trên \(\left[-1;8\right]\)
\(f\left(-1\right)=3\) ; \(f\left(-\frac{b}{2a}\right)=f\left(\frac{5}{2}\right)=-\frac{37}{4}\); \(f\left(8\right)=21\)
\(\Rightarrow f\left(x\right)_{min}=f\left(\frac{5}{2}\right)=-\frac{37}{4}\)
\(f\left(x\right)_{max}=f\left(8\right)=21\)
\(f\left(x\right)=\left(2-x\right)\left(x+3\right)\le\dfrac{1}{4}\left(2-x+x+3\right)^2=\dfrac{25}{4}\)
\(f\left(x\right)_{max}=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)
c) \(h\left(x\right)=\left(x+1\right)^2+\left(\dfrac{x^2+2x+2}{x+1}\right)^2=\left(x+1\right)^2+\left(x+1+\dfrac{1}{x+1}\right)^2=2\left(x+1\right)^2+\dfrac{1}{\left(x+1\right)^2}+2\ge_{AM-GM}2\sqrt{2}+2\).
Đẳng thức xảy ra khi \(2\left(x+1\right)^2=\dfrac{1}{\left(x+1\right)^2}\Leftrightarrow x=\pm\sqrt{\dfrac{1}{2}}-1\).
b) \(g\left(x\right)=\dfrac{\left(x+2\right)\left(x+3\right)}{x}=\dfrac{x^2+5x+6}{x}=\left(x+\dfrac{6}{x}\right)+5\ge_{AM-GM}2\sqrt{6}+5\).
Đẳng thức xảy ra khi x = \(\sqrt{6}\).
Lời giải:
Để hàm đồng biến trên $R$ thì:
$m+1>0$
$\Leftrightarrow m>-1$
Mà $m$ nguyên và $m\in [-3;3]$ nên $m\in\left\{0;1;2;3\right\}$
Vậy có 4 giá trị thỏa mãn.