已知函数f(x)=3x-4的值域为[-10,5]-搜答案-拍题作业网

x=0时,f(x)单调递减,f(0)=1>0、f(1)=1/3-1=-2/3

(1)f(5)=-5+2=-3f(f(5))=f(-3)=-3+4=1f(f(f(5))=f(1)=1^2-1*2=-1(2)

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这个函数的图象实际上是函数y=x^2-4x+3在x轴下方的图象以x轴对称翻折到x轴上方而已y=x^2-4x+3的对称轴为x=2与x轴的交点为(1.0)(3.0)一眼就可以看出所以（负无穷.1）递减[1

f(x)=ax+bf(f(x))=a(ax+b)+b=a^2x+ab+b=4x-3a^2=4a=2a=-2ab+b=-3a=2b=-1f(x)=2x-1a=-2b=3f(x)=-2x+3

f(1/4)=log2(1/4)=log2(2^(-2)）=-2f(-2)=3^(-2)=1/9所以为1/9

设f(x)=kxb代入内函数中,即f【f(x)】=f(kxb)将kxb看作自变量,代入外函数中,即f(kxb)=(kxb)kb展开得：k2xkbb=4x-1左右两边系数相同,即k2=4,kbb=-1所

f(x+a)=(x-2)^4-16令x=x-a代入得f(x)=(x-2-a)^4-16f(a)=0f(f(a))=f(0)=(2+a)^4-16=3(2+a)^4=194次方对吗?

定义域：sin(x+π/4)≠0,则x≠kπ-π/4（k为—∞到∞上的整数）f(x)=cos2x/sin(x+π/4)=4/33cos2x=4sin(x+π/4)3(cos²x-sin

1/9分别按照函数定义带入f(1/4)=log2(1/4)=-2f(f(1/4))=f(-2)=3^(-2)=1/9

f（1）=5f（-3）=-3f（a+1）=（a+1）（a+5）

1.f(π/9)=tan(π/3+π/4)=(tanπ/3+tanπ/4)/(1-tanπ/3tanπ/4)=(√3+1)/(1-√3)=-√3-22.∵f(a/3+π/4)=2∴tan(a+3π/4

f(x)=ax/(2x+3)f[f(x)]=a[ax/(2x+3)]/[2ax/(2x+3)+3]=xa[ax/(2x+3)]/[2ax/(2x+3)+3]=x左边上下乘2x+3a^2x/(2ax+6

sin2x=-cos(2x+π/2)=-1+2sin²(x+π/4)sin(x+π/4)=√（sin2x+1）/2f(x)=cos2x/(√（sin2x+1）/2)=4/3sin2x=-1或

f(x)=∫f'(x)dx=x³/3+2x²+cf(-3)=-9+18+c=10c=1f(x)=x³/3+2x²+1

f=ax+bf[f(x)]=a(ax+b)+b=a²x+ab+ba²=4ab+b=3a=2b=1a=-2b=-3

f(π/3-x)=cos(π/3-x)/cos(π/6-(π/3-x))=cos(π/3-x)/cos(x-π/6)=cos(π/3-x)/cos(π/6-x)注：cosx=cos(-x)所以：f(x

f(1/4)=log2(1/4)=-2f(1/4)小于零所以f[f(1/4)]=3^-2=1/9

2f(x)+f(-x)=2x-34f(x)+2f(-x)=4x-62f(-x)+f(x)=-2x-33f(x)=6x-3f(x)=2x-1

∵y=f(x)的导函数为f`(x)=3x^2-6x∴f（x）=x^3-3x^2C（C为常数）又∵f（0）=4∴C=4∴f（x）=x^3-3x^24令f'(x)＜0,解得0＜x＜2∴f（x）的单调减区间