Problem...

[Jordann]

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]

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]

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]

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) = e^x-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.