Problem...

Jordann

New member
Joined
Apr 13, 2019
Messages
2
Hello everybody, i have to solve the following problem:
Its is given that f(x) >= e^(x-1)+lnx+x^2 for x>0 and f(1) = 2
I have to find the tangent of f in A(1,2) which is: y-f(1)=f ' (1)(x-1)
I have trouble finding f ' (1)
I tried finding it through extremum but didnt manage anything. Any help would be appreciated....
 
Hello everybody, i have to solve the following problem:
Its is given that f(x) >= e^(x-1)+lnx+x^2 for x>0 and f(1) = 2
I have to find the tangent of f in A(1,2) which is: y-f(1)=f ' (1)(x-1)
I have trouble finding f ' (1)
If f(x)ex1+log(x)+x2\displaystyle f(x)\ge e^{x-1}+\log(x)+x^2 forx>0\displaystyle x>0 and f(1)=2\displaystyle f(1) = 2 then
f(x)=ex1+1x+2x\displaystyle f'(x)=e^{x-1}+\frac{1}{x}+2x therefore f(1)= ?\displaystyle f'(1)=~?
 
I mean the equation.
If f(x)ex1+log(x)+x2\displaystyle f(x)\ge e^{x-1}+\log(x)+x^2 forx>0\displaystyle x>0 and f(1)=2\displaystyle f(1) = 2 then
f(x)=ex1+1x+2x\displaystyle f'(x)=e^{x-1}+\frac{1}{x}+2x therefore f(1)= ?\displaystyle f'(1)=~?
You cant simply make an equality from an unequality. Seems like a new function h(x) = ex-1+log(x)+x2
and then h(x)>=0 or h(x) >= h(1) so h(1) is total minimum and from ferma theorem h'(1) = 0. That is the only probable solution i think.
 
Top