1. show that g(x) = x^2 / (1+x^2) is an increasing function for all x > 0
2. show that g(x) = x [(1+x^2)^1/2 ] is a strictly increasing function for all real values of x
3. show that g(x) = 5e^(-2x) is a strictly decreasing function for all real values of x
m2 differentiation
2012-11-17 11:17 pm
回答 (2)
2012-11-18 12:34 am
✔ 最佳答案
1.g'(x)=d/dx(x^2/(1+x^2))=((1+x^2)(2x)-(x^2)(2x))/(1+x^2)^2
=2x/(1+x^2)^2
when x>0,g'(x)>0
g(x)is an increasing function for all x>0
2.g'(x)=d/dx(x(1+x^2)^(1/2))
=x(1/2(1+x^2)^(-1/2))(2x)+(1+x^2)^(1/2)
=x^2(1+x^2)^(-1/2)+(1+x^2)^(1/2)>0
g(x)is a strictly increasing function for all real values of x.
3.g'(x)=d/dx5e^(-2x)
=5e^(-2x)(-2)
=-10e^(-2x)<0
g(x)is a strictly decreasing function for all real values of x.
參考: me
2012-11-18 12:14 am
1) g(x) = 1 - 1/(1 + x^2)
g'(x) = 2x/(1 + x^2)^2 > 0 for all positive x and hence g is strictly increasing
2) g'(x) = (1 + x^2)^1/2 + x^2/(1 + x^2)^1/2 which is positive for all real x, so g is strictly increasing
3) g'(x) = -10e^-2x which is negative for all real x, so g is strictly decreasing.
g'(x) = 2x/(1 + x^2)^2 > 0 for all positive x and hence g is strictly increasing
2) g'(x) = (1 + x^2)^1/2 + x^2/(1 + x^2)^1/2 which is positive for all real x, so g is strictly increasing
3) g'(x) = -10e^-2x which is negative for all real x, so g is strictly decreasing.
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