x=e^t
therefore
g(t)=( 1/2)(e^t+e^(-t))
h(t)=( 1/2)(e^t-e^(-t))
f(t)=g(t)+h(t)=e^t
f(x)=e^x
Let f (x) = e x-1-x-x 2/2.
F(0)= 1- 1-0-0=0
F'(x)=e^x- 1-x
F'(x)>0 to x>0 holds.
Because F'(0)= 1- 1-0=0.
The next requirement is that f'' (x) = e x- 1 > 0 to x>0 holds.
Obviously, because e x is increasing, when x=0, the function value is 1, x >;; E x of 0 must be greater than 1
So F''(x) is constant and increasing when x >: 0, so F'(x) >; 0
So 0 in F(x)>x >; 0 Shangheng is established.
So f (x) >1+x+x 2/2 in x > 0 Shangheng is established.
What are the characteristics of Montessori mathematics?
1. Gamification of teaching AIDS: This teaching aid is designed wi