《章末整合》函數(shù)PPT
第一部分內(nèi)容：探究學(xué)習(xí)
題型一、分段函數(shù)的應(yīng)用
例1已知函數(shù)f(x)={■("-" x^2+2x"," x>0"," @0"," x=0"," @x^2+mx"," x<0)┤是奇函數(shù).
(1)求實(shí)數(shù)m的值;
(2)若函數(shù)f(x)在區(qū)間[-1,a-2]上單調(diào)遞增,求實(shí)數(shù)a的取值范圍.
解:(1)設(shè)x<0,則-x>0,
∴f(-x)=-(-x)2+2(-x)=-x2-2x.
又f(x)為奇函數(shù),∴f(-x)=-f(x).
∴當(dāng)x<0時(shí),f(x)=x2+2x=x2+mx,∴m=2.
(2)要使f(x)在[-1,a-2]上單調(diào)遞增,結(jié)合f(x)的圖像(圖像略)知
{■(a"-" 2>"-" 1"," @a"-" 2≤1"," )┤∴1<a≤3,故實(shí)數(shù)a的取值范圍是(1,3].
方法技巧已知函數(shù)的奇偶性求參數(shù)值,可利用定義或特殊值來求解,本題也可用f(-1)=-f(1)求出m的值,再檢驗(yàn)即可.另外,分段函數(shù)的各段的單調(diào)性可分別判斷,但對于跨段的單調(diào)性問題要注意在分段端點(diǎn)處的銜接.
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題型二、函數(shù)單調(diào)性、奇偶性的綜合應(yīng)用
例2已知函數(shù)f(x)=ax+ (x≠0,常數(shù)a∈R).
(1)討論函數(shù)f(x)的奇偶性,并說明理由;
(2)若函數(shù)f(x)在x∈[3,+∞)上為增函數(shù),求a的取值范圍.
解:(1)f(x)的定義域?yàn)閧x|x≠0},其定義域關(guān)于原點(diǎn)對稱.
當(dāng)a=0時(shí),f(x)=1/x^2 ,對任意x∈(-∞,0)∪(0,+∞),
f(-x)=1/("(-" x")" ^2 )=1/x^2 =f(x),∴f(x)為偶函數(shù).
當(dāng)a≠0時(shí),f(x)=ax+1/x^2 (a≠0,x≠0),取x=±1,
得f(-1)+f(1)=2≠0,f(-1)-f(1)=-2a≠0,
∴函數(shù)f(x)是非奇非偶函數(shù).
綜上所述,當(dāng)a=0時(shí),f(x)是偶函數(shù);
當(dāng)a≠0時(shí),f(x)是非奇非偶函數(shù).
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題型三、二次函數(shù)的最值(值域)
例3已知函數(shù)f(x)=x2+2ax+2.
(1)當(dāng)a=-1時(shí),求函數(shù)f(x)在區(qū)間[-5,5]上的最大值和最小值;
(2)用a表示出函數(shù)f(x)在區(qū)間[-5,5]上的最值.
分析:將原函數(shù)先配方,對于第(2)問還要結(jié)合圖像進(jìn)行分類討論.
解:(1)當(dāng)a=-1時(shí),f(x)=x2-2x+2=(x-1)2+1,
因?yàn)?∈[-5,5],故當(dāng)x=1時(shí),f(x)取得最小值,f(x)min=f(1)=1;
當(dāng)x=-5時(shí),f(x)取得最大值,f(x)max=f(-5)=(-5-1)2+1=37.
(2)函數(shù)f(x)=x2+2ax+2=(x+a)2+2-a2的圖像開口向上,對稱軸為x=-a.
當(dāng)-a≤-5,即a≥5時(shí),函數(shù)在區(qū)間[-5,5]上是增函數(shù),所以f(x)max=f(5)=27+10a,f(x)min=f(-5)=27-10a;
當(dāng)-5<-a≤0,即0≤a<5時(shí),函數(shù)圖像如圖①所示,由圖像可得f(x)min=f(-a)=2-a2,f(x)max=f(5)=27+10a;
當(dāng)0<-a<5,即-5<a<0時(shí),函數(shù)圖像如圖②所示,由圖像可得f(x)max=f(-5)=27-10a,f(x)min=f(-a)=2-a2;
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