高等数学题目求解
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(2)
f(x)=cosx/√(1-x^2)
f(-x) = cosx/√(1-x^2)=f(x)
f(x) 偶函数
(3)
lim(x->3) (x^2-5x+6)/(x^2-9)
=lim(x->3) (x-2)(x-3)/[(x-3)(x+3)]
=lim(x->3) (x-2)/(x+3)
=(3-2)/(3+3)
=1/6
(2)
y=2x
lim(x->+无穷) [1 - 1/(2x)]^x
=lim(y->+无穷) [1 - 1/y]^(y/2)
=e^(-1/2)
(4)
f(x)
=(1/2)x^2-3x+2
=(1/2)[x^2-6x]+2
=(1/2)[x^2-6x+9]+2 -9/2
=(1/2)(x-3)^2 - 5/2
min f(x)=f(3) =-5/2
f(-1) =(1/2)(16) -5/2 = 11/2
f(4)=(1/2)(1) -5/2 = -2
max f(x) = f(-1)=11/2
(5)
(1)
x^4
=x^2.(1+x^2) -x^2
=x^2.(1+x^2) -(1+x^2) +1
∫ x^4/(1+x^2) dx
=∫ [ x^2-1 + 1/(1+x^2)] dx
=(1/3)x^3 -x +arctanx +C
(2)
∫ (1+lnx)^3/x dx
=∫ (1+lnx)^3 d(1+lnx)
=(1/4)(1+lnx)^4 +C
(6)
y'-2xy=0
dy/dx = 2xy
∫dy/y =∫ 2x dx
ln|y| = x^2 +C'
y= C.e^(x^2)
f(x)=cosx/√(1-x^2)
f(-x) = cosx/√(1-x^2)=f(x)
f(x) 偶函数
(3)
lim(x->3) (x^2-5x+6)/(x^2-9)
=lim(x->3) (x-2)(x-3)/[(x-3)(x+3)]
=lim(x->3) (x-2)/(x+3)
=(3-2)/(3+3)
=1/6
(2)
y=2x
lim(x->+无穷) [1 - 1/(2x)]^x
=lim(y->+无穷) [1 - 1/y]^(y/2)
=e^(-1/2)
(4)
f(x)
=(1/2)x^2-3x+2
=(1/2)[x^2-6x]+2
=(1/2)[x^2-6x+9]+2 -9/2
=(1/2)(x-3)^2 - 5/2
min f(x)=f(3) =-5/2
f(-1) =(1/2)(16) -5/2 = 11/2
f(4)=(1/2)(1) -5/2 = -2
max f(x) = f(-1)=11/2
(5)
(1)
x^4
=x^2.(1+x^2) -x^2
=x^2.(1+x^2) -(1+x^2) +1
∫ x^4/(1+x^2) dx
=∫ [ x^2-1 + 1/(1+x^2)] dx
=(1/3)x^3 -x +arctanx +C
(2)
∫ (1+lnx)^3/x dx
=∫ (1+lnx)^3 d(1+lnx)
=(1/4)(1+lnx)^4 +C
(6)
y'-2xy=0
dy/dx = 2xy
∫dy/y =∫ 2x dx
ln|y| = x^2 +C'
y= C.e^(x^2)
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