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....
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,979
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 \(\displaystyle f(x)\ge e^{x-1}+\log(x)+x^2\) for\(\displaystyle x>0\) and \(\displaystyle f(1) = 2\) then
\(\displaystyle f'(x)=e^{x-1}+\frac{1}{x}+2x\) therefore \(\displaystyle f'(1)=~?\)
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,351
I have to find the tangent of f
What do you mean by 'tangent'?? Do you meant the slope of the tangent line or the equation of the tangent line or something else?
 

Jordann

New member
Joined
Apr 13, 2019
Messages
2
I mean the equation.
If \(\displaystyle f(x)\ge e^{x-1}+\log(x)+x^2\) for\(\displaystyle x>0\) and \(\displaystyle f(1) = 2\) then
\(\displaystyle f'(x)=e^{x-1}+\frac{1}{x}+2x\) therefore \(\displaystyle f'(1)=~?\)
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.
 
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