帮求数学第2，3题， 下图
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(2)
f(x)
=bx+2 ; x≤0
=a+ln(x+1) ; x>0
f(0-)=f(0) = 2
lim(x->0+) f(x) = lim(x->0+) [ a+ln(x+1) ] =a
f(0-)=f(0+) =>a=2
f'(0-) = b
f'(0+)
= lim(x->0+) [ 2+ln(x+1) - f(0) ]/x
= lim(x->0+) ln(x+1)/x
=1
=> b=1
(a,b) = (2,1)
(3)
y= 2^(cos3x) + (cos3x)^2
y' =-2(sin3x)(ln2). 2^(cos3x) -6(cos3x).(sin3x)
y'|x=0 =0
f(x)
=bx+2 ; x≤0
=a+ln(x+1) ; x>0
f(0-)=f(0) = 2
lim(x->0+) f(x) = lim(x->0+) [ a+ln(x+1) ] =a
f(0-)=f(0+) =>a=2
f'(0-) = b
f'(0+)
= lim(x->0+) [ 2+ln(x+1) - f(0) ]/x
= lim(x->0+) ln(x+1)/x
=1
=> b=1
(a,b) = (2,1)
(3)
y= 2^(cos3x) + (cos3x)^2
y' =-2(sin3x)(ln2). 2^(cos3x) -6(cos3x).(sin3x)
y'|x=0 =0
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请问第三题怎么做 谢谢
追答
请问第三题
(3)
y= 2^(cos3x) + (cos3x)^2
y' =-3(sin3x)(ln2). 2^(cos3x) -6(cos3x).(sin3x)
y'|x=0 =0
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