Inverse function y=lnx y'= 1/x

Tangency means that there are intersections.

So there's an equation.

k= 1/x

y=kx+ 1

y=lnx

y=2,x=e^2,k=e^-2

2. What do you mean by * * *? Is it a crossroads?

When x=0, f (0) = 1, MX 2 = 0.

x-& gt； Lim e x/MX 2 = lim e x/2m =∞ at infinity; Function f (x) > MX 2;

And because both derivatives are greater than 0, they are in x >; There is no intersection at 0.

Maybe my method is wrong. . .

e^b-e^a/(b-a)=e^a(e^(b-a)- 1)/(b-a)

e^b+e^a/2=e^a(e^(b-a)+ 1)/2

Let x = b-a; x & gt0

e^a* e^x- 1 / x

e^a* e^x+ 1 / 2

Extraction ratio

g(x)=2(e^x- 1)/[x(e^x+ 1)]x & gt； 0

When x->; 0，g(x)= 1； The definition of derivative can also be directly derived.

g'(x)=2[(e^x)[x(e^x+ 1)]-(e^x- 1)(e^x+ 1+xe^x)]/[x(e^x+ 1)]^2

Molecular partial expansion = (ex) [x (ex+1)]-(ex-1) (ex+1+xex) t = ex >; 1

txt+tx- {txt+tt+t-tx-t- 1}

= 1-tt= 1-e^2x；

When x>0, the molecule

So g' (x) < 0, g (x) is a decreasing function, and g(x)

e^b-e^a/(b-a)/x & lt； e^b+e^a/2