# Problem...

#### Jordann

##### New member
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
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 $$\displaystyle f'(1)=~?$$

• topsquark

#### Jomo

##### Elite Member
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
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.