2个回答
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1. lim(x→0)(1+2/x)^(3x+1)
= lim(x→0){[(1+2/x)^(x/2)]^6}* (1+2/x)
= e^6;
2. dy/dx = f'[(3x-2)/(3x+2)]*[12/(3x+2)^2],
于是,
dy(0)/dx = f'(-1)*(12/2^2) = 3f'(-1) = 3π/2;
3. 先算
f'-(0) = lim(x→0-)[f(x) - f(0)]/x
= lim(x→0-)(x^2)/x = 0,
f'+(0) = lim(x→0+)[f(x) - f(0)]/x
= lim(x→0-){[x(e^x)] - 0}/x = 1,
知函数在 x = 0 不可导;所以
f'(x) = 2x, x < 0,
= (x+1)e^x,x > 0;
4. ∫[0, 4](1/√x)f(√x)dx
= 2∫[0, 4]f(√x)d(√x)
= 2∫[0, 2]f(t)dt (令 t = √x)
= 2*(2^4)/2 = ……
5. ∫[3, 5]f(t)dt
= 2∫[1, 2]f(2x+1)dx (令 t = 2x+1)
= 2∫[1, 2]x(e^x)dx
= 2[(x+1)e^x]|[1, 2]
= ……
= lim(x→0){[(1+2/x)^(x/2)]^6}* (1+2/x)
= e^6;
2. dy/dx = f'[(3x-2)/(3x+2)]*[12/(3x+2)^2],
于是,
dy(0)/dx = f'(-1)*(12/2^2) = 3f'(-1) = 3π/2;
3. 先算
f'-(0) = lim(x→0-)[f(x) - f(0)]/x
= lim(x→0-)(x^2)/x = 0,
f'+(0) = lim(x→0+)[f(x) - f(0)]/x
= lim(x→0-){[x(e^x)] - 0}/x = 1,
知函数在 x = 0 不可导;所以
f'(x) = 2x, x < 0,
= (x+1)e^x,x > 0;
4. ∫[0, 4](1/√x)f(√x)dx
= 2∫[0, 4]f(√x)d(√x)
= 2∫[0, 2]f(t)dt (令 t = √x)
= 2*(2^4)/2 = ……
5. ∫[3, 5]f(t)dt
= 2∫[1, 2]f(2x+1)dx (令 t = 2x+1)
= 2∫[1, 2]x(e^x)dx
= 2[(x+1)e^x]|[1, 2]
= ……
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