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判斷極值的正負(fù)(2021新高考Ⅱ?qū)?shù))

2022-10-15 10:45 作者:數(shù)學(xué)老頑童  | 我要投稿


解:(1)求導(dǎo)得

%5Cbegin%7Baligned%7D%0A%09f'%5Cleft(%20x%20%5Cright)%20%26%3D1%5Ccdot%20%5Cmathrm%7Be%7D%5Ex%2B%5Cleft(%20x-1%20%5Cright)%20%5Ccdot%20%5Cmathrm%7Be%7D%5Ex-2ax%5C%5C%0A%09%26%3Dx%5Cleft(%20%5Cmathrm%7Be%7D%5Ex-2a%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D

1.%5Ccolor%7Bred%7D%7Ba%5Cleqslant%200%7D,

f'%5Cleft(%20x%20%5Cright)%20%3D0,得x%3D0,

當(dāng)x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%20-%5Cinfty%20%2C0%20%5Cright)%20%7D,f'%5Cleft(%20x%20%5Cright)%20%3C0,f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Csearrow%20%7D;

當(dāng)x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%200%2C%2B%5Cinfty%20%5Cright)%20%7D,f'%5Cleft(%20x%20%5Cright)%20%3E0,f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

2.%5Ccolor%7Bred%7D%7Ba%3E0%7D,

f'%5Cleft(%20x%20%5Cright)%20%3D0,得x%3D0,或x%3D%5Cln%20%202a.

2.1%5Cln%20%202a%3E0,即%5Ccolor%7Bred%7D%7Ba%3E%5Cfrac%7B1%7D%7B2%7D%7D,

當(dāng)x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%20-%5Cinfty%20%2C0%20%5Cright)%20%7D,f'%5Cleft(%20x%20%5Cright)%20%3E0,f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

當(dāng)x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%200%2C%5Cln%20%202a%20%5Cright)%20%7D,f'%5Cleft(%20x%20%5Cright)%20%3C0,f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Csearrow%20%7D

當(dāng)x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%20%5Cln%20%202a%2C%2B%5Cinfty%20%5Cright)%20%7D,f'%5Cleft(%20x%20%5Cright)%20%3E0,f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

2.2%5Cln%20%202a%3D0,即%5Ccolor%7Bred%7D%7Ba%3D%5Cfrac%7B1%7D%7B2%7D%7D,

f'%5Cleft(%20x%20%5Cright)%20%5Cgeqslant%200f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D.

2.3%5Cln%20%202a%3C0,即%5Ccolor%7Bred%7D%7B0%3Ca%3C%5Cfrac%7B1%7D%7B2%7D%7D,

當(dāng)x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%20-%5Cinfty%20%2C%5Cln%20%202a%20%5Cright)%20%7D,f'%5Cleft(%20x%20%5Cright)%20%3E0,f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

當(dāng)x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%20%5Cln%20%202a%2C0%20%5Cright)%20%7D,f'%5Cleft(%20x%20%5Cright)%20%3C0,f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Csearrow%20%7D

當(dāng)x%5Cin%20%5Ccolor%7Bred%7D%7B%5Cleft(%200%2C%2B%5Cinfty%20%5Cright)%20%7D,f'%5Cleft(%20x%20%5Cright)%20%3E0,f%5Cleft(%20x%20%5Cright)%20%5Ccolor%7Bred%7D%7B%5Cnearrow%20%7D

(2)f%5Cleft(%200%20%5Cright)%20%3Db-1,

f%5Cleft(%20%5Cln%20%202a%20%5Cright)%20%3D%5Cleft(%20%5Ccolor%7Bred%7D%7B2-%5Cln%20%202a%7D%20%5Cright)%20%5Ccdot%20a%5Ccdot%20%5Ccolor%7Bred%7D%7B%5Cln%20%202a%7D%2B%5Ccolor%7Bred%7D%7Bb-2a%7D

若選①,由(1)知:

f%5Cleft(%20x%20%5Cright)%20的極大值為f%5Cleft(%200%20%5Cright)%20,極小值為f%5Cleft(%20%5Cln%20%202a%20%5Cright)%20

因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=%5Cfrac%7B1%7D%7B2%7D%3Ca%5Cleqslant%20%5Cfrac%7B%5Cmathrm%7Be%7D%5E2%7D%7B2%7D" alt="%5Cfrac%7B1%7D%7B2%7D%3Ca%5Cleqslant%20%5Cfrac%7B%5Cmathrm%7Be%7D%5E2%7D%7B2%7D">,

所以%5Ccolor%7Bred%7D%7B0%3C%5Cln%20%202a%7D%3C2

所以%5Ccolor%7Bred%7D%7B0%3C2-%5Cln%20%202a%7D%3C2;

因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=b%3E2a" alt="b%3E2a">,所以%5Ccolor%7Bred%7D%7Bb-2a%3E0%7D

所以%5Ccolor%7Bred%7D%7Bf%5Cleft(%20%5Cln%20%202a%20%5Cright)%20%3E0%7D,

自然易知f%5Cleft(%200%20%5Cright)%20%3E0,

f%5Cleft(%20x%20%5Cright)%20%5Cleft%5B%200%2C%2B%5Cinfty%20%5Cright)%20無(wú)零點(diǎn);

又因f%5Cleft(%20%5Ccolor%7Bred%7D%7B-%5Csqrt%7B%5Cfrac%7Bb%7D%7Ba%7D%7D%7D%20%5Cright)%20%3C0,

%5Cexists%20x_0%5Cin%20%5Cleft(%20-%5Csqrt%7B%5Cfrac%7Bb%7D%7Ba%7D%7D%2C0%20%5Cright)%20,使f%5Cleft(%20x_0%20%5Cright)%20%3D0,

證畢.

若選②,由(1)知:

f%5Cleft(%20x%20%5Cright)%20的極大值為f%5Cleft(%20%5Cln%20%202a%20%5Cright)%20,極小值為f%5Cleft(%200%20%5Cright)%20

因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=0%3Ca%3C%5Cfrac%7B1%7D%7B2%7D" alt="0%3Ca%3C%5Cfrac%7B1%7D%7B2%7D">,

所以0%3C2a%3C1

所以%5Ccolor%7Bred%7D%7B%5Cln%20%202a%3C0%7D,

所以%5Ccolor%7Bred%7D%7B2-%5Cln%20%202a%7D%3E2%5Ccolor%7Bred%7D%7B%3E0%7D

因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=b%5Cleqslant%202a" alt="b%5Cleqslant%202a">,

所以%5Ccolor%7Bred%7D%7Bb-2a%5Cleqslant%200%7D,

所以%5Ccolor%7Bred%7D%7Bf%5Cleft(%20%5Cln%20%202a%20%5Cright)%20%3C0%7D,

自然易知f%5Cleft(%200%5Cright)%20%3C0

f%5Cleft(%20x%20%5Cright)%20%5Cleft(%20-%5Cinfty%20%2C0%20%5Cright%5D%20無(wú)零點(diǎn),

又因?yàn)?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=f%5Cleft(%20%5Ccolor%7Bred%7D%7B2-%5Cfrac%7Bb%7D%7B%5Cmathrm%7Be%7D%5E%7Ba%2B1%7D%7D%7D%20%5Cright)%20%3E0" alt="f%5Cleft(%20%5Ccolor%7Bred%7D%7B2-%5Cfrac%7Bb%7D%7B%5Cmathrm%7Be%7D%5E%7Ba%2B1%7D%7D%7D%20%5Cright)%20%3E0">,

%5Cexists%20x_0%5Cin%20%5Cleft(%200%2C2-%5Cfrac%7Bb%7D%7B%5Cmathrm%7Be%7D%5E%7Ba%2B1%7D%7D%20%5Cright)%20,

使得f%5Cleft(%20x_0%20%5Cright)%20%3D0,證畢.


判斷極值的正負(fù)(2021新高考Ⅱ?qū)?shù))的評(píng)論 (共 條)

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