前言
典例剖析
(1).求\(f(x)\)的单调区间;
分析:化简为正弦型或者余弦型,
\(f(x)=\cfrac{\sqrt{3}}{3}[cos(2x+\cfrac{\pi}{6})+4sinxcosx]+1\),
\(=\cfrac{\sqrt{3}}{3}[(cos2x\cdot cos\cfrac{\pi}{6}-sin2x\cdot sin\cfrac{\pi}{6})+2sin2x]+1\)
\(=\cfrac{\sqrt{3}}{3}(\cfrac{\sqrt{3}}{2}cos2x-\cfrac{1}{2}sin2x+2sin2x)+1\)
\(=\cfrac{\sqrt{3}}{3}(\cfrac{\sqrt{3}}{2}cos2x+\cfrac{3}{2}sin2x)+1\)
\(=\cfrac{\sqrt{3}}{2}sin2x+\cfrac{1}{2}cos2x+1\)
即函数\(f(x)=sin(2x+\cfrac{\pi}{6})+1\),[化简到此结束]
令\(2k\pi-\cfrac{\pi}{2}\leqslant 2x+\cfrac{\pi}{6}\leqslant 2k\pi+\cfrac{\pi}{2}\),\(k\in Z\),
解得单调递增区间为\([k\pi-\cfrac{\pi}{3},k\pi+\cfrac{\pi}{6}]\),\((k\in Z)\);
令\(2k\pi+\cfrac{\pi}{2}\leqslant 2x+\cfrac{\pi}{6}\leqslant 2k\pi+\cfrac{3\pi}{2}\),\(k\in Z\),
解得单调递减区间为\([k\pi+\cfrac{\pi}{6},k\pi+\cfrac{2\pi}{3}]\),\((k\in Z)\);
注意
:求\(f(x)=Asin(\omega x+\phi)+k\)型函数的单调区间,为减少错误,一般需要将\(\omega\)转化为\(\omega>0\);
(2).令\(g(x)=af(x)+b\),若函数\(g(x)\)在区间\([-\cfrac{\pi}{6},\cfrac{\pi}{4}]\)上的值域为\([-1,1]\),求\(a+b\)的值;
提示:当\(x\in [-\cfrac{\pi}{6},\cfrac{\pi}{4}]\)时,\(sin(2x+\cfrac{\pi}{6})\in [-\cfrac{1}{2},1]\),则\(f(x)\in [\cfrac{1}{2},2]\),
以下针对一次项系数\(a\)分类讨论如下:
①当\(a>0\)时,由\(2a+b=1\)和\(\cfrac{1}{2}a+b=-1\),解得\(a=\cfrac{4}{3}\),\(b=-\cfrac{5}{3}\),故\(a+b=-\cfrac{1}{3}\);
②当\(a<0\)时,由\(2a+b=-1\)和\(\cfrac{1}{2}a+b=1\),解得\(a=-\cfrac{4}{3}\),\(b=\cfrac{5}{3}\),故\(a+b=\cfrac{1}{3}\);
故\(a+b\)的值为\(\cfrac{1}{3}\)或\(-\cfrac{1}{3}\);
(1).求函数的定义域;
分析:由函数解析式可知,需要让\(tanx\)有意义,故定义域为\(\{x\mid x\neq k\pi+\cfrac{\pi}{2},k\in Z\}\)
(2).试讨论\(f(x)\)在区间\([-\cfrac{\pi}{4},\cfrac{\pi}{4}]\)上的单调性。
分析:先将所给函数化简为正弦型或者余弦型,
\(f(x)=4tan\cdot cosx(cosx\cdot \cfrac{1}{2}+sinx\cdot \cfrac{\sqrt{3}}{2})-\sqrt{3}\)
\(=4sinx(cosx\cdot \cfrac{1}{2}+sinx\cdot \cfrac{\sqrt{3}}{2})-\sqrt{3}\)
\(=2sinxcosx+2\sqrt{3}sin^2x-\sqrt{3}\)
\(=sin2x+\sqrt{3}(1-cos2x)-\sqrt{3}\)
\(=sin2x-\sqrt{3}cos2x\)
\(=2sin(2x-\cfrac{\pi}{3})\)
法1:先求解函数在\(x\in R\)上的单调区间,
令\(2k\pi-\cfrac{\pi}{2}\leq 2x-\cfrac{\pi}{3}\leq 2k\pi+\cfrac{\pi}{2}(k\in Z)\),
得到单调递增区间为\([k\pi-\cfrac{\pi}{12},k\pi+\cfrac{5\pi}{12}](k\in Z)\),
又因为\(x\in [-\cfrac{\pi}{4},\cfrac{\pi}{4}]\),
[说明:求得的单调递增区间和给定区间求交集,即为所求的单调递增区间;剩余的即为单调递减区间]
然后给\(k\)赋值,令\(k=0\),
得到函数在区间\([-\cfrac{\pi}{12},\cfrac{\pi}{4}]\)上单调递增,在区间\([-\cfrac{\pi}{4},-\cfrac{\pi}{12}]\)上单调递减。
法2:由\(-\cfrac{\pi}{4}\leq x\leq \cfrac{\pi}{4}\),求得\(-\cfrac{5\pi}{6}\leq 2x-\cfrac{\pi}{3}\leq \cfrac{\pi}{6}\),结合横轴为\(2x-\cfrac{\pi}{3}\)的图像可知,
当\(-\cfrac{5\pi}{6}\leq 2x-\cfrac{\pi}{3}\leq -\cfrac{\pi}{2}\)时,求得函数在区间\([-\cfrac{\pi}{4},-\cfrac{\pi}{12}]\)单调递减;
当\(-\cfrac{\pi}{2}\leq 2x-\cfrac{\pi}{3}\leq \cfrac{\pi}{6}\)时,求得函数在区间\([-\cfrac{\pi}{12},\cfrac{\pi}{4}]\)单调递增;
来源:https://www.cnblogs.com/wanghai0666/p/12015078.html