作业帮导数数以及三角函数题
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导数数以及三角函数题
导数数以及三角函数题
导数数以及三角函数题
(11)
f(x)=acoswx-sinwx central symmetric about M( π/3,0), w>0
min f(x) = f(π/6)
1 possible value of a+w
Solution:
f(x)=acoswx-sinwx
f'(x) = -awsinwx- wcoswx
f'(π/6) = 0
-asin(πw/6)- cos(πw/6) =0 (1)
f(x)=acoswx-sinwx central symmetric about M(π/3,0)
f(0) = a
f(2π/3) = -f(0)
f(2π/3) = acos(2πw/3)-sin(2πw/3) =-a
a(1+ cos(2πw/3)) = sin(2πw/3)
a = tan(πw/3) (2)
sub (2) into (1)
sin(πw/6)sin(πw/3)+ cos(πw/6)cos(πw/3) =0
cos(πw/6) = 0
πw/6 = π/2, 3π/2,...
w = 3, 9, .
from (1)
-awsin(πw/6)- wcos(πw/6) =0
w=3
a=0
1 possible value a+w = 0+3 =3
(12)
y=f(x)=a^x ,( a>0, a≠1), symmetric about y=x
g(x) = f(x)[f(x)+f(2)-1]
y=g(x) is increasing [1/2,2]
Find: range of a
Solution:
y=f(x)=a^x ,( a>0, a≠1), symmetric about y=x
我不懂,不过我觉得题目有点问题.
y=f(x)=a^x ,( a>0, a≠1), 不可能对称于 y=x
因为
(0,1) 属于y=f(x)
(0,1)的相关对,y=x, 的称点是(1,0)
但是(1,0)不属于y=f(x)?