I'm having trouble finding the extremas of...
I'm having trouble finding the crit # of this. Please show me how to do it...
f(x)= e^x/1-x+x^2
This is $\displaystyle f(x) = \frac{e^x}{1 - x + x^2}$ I presume? Please use parenthesis!
$\displaystyle f^{\prime}(x) = \frac{e^x(1 - x + x^2) - e^x(-1 + 2x)}{(1 - x + x^2)^2}$
$\displaystyle f^{\prime}(x) = \frac{e^x(2 - 3x + x^2)}{(1 - x + x^2)^2}$
You will have critical numbers where $\displaystyle f^{\prime}(x) = 0$, so solve
$\displaystyle \frac{e^x(2 - 3x + x^2)}{(1 - x + x^2)^2} = 0$
$\displaystyle e^x(x^2 - 3x + 2) = 0$
$\displaystyle e^x(x - 1)(x - 2) = 0$
So either
$\displaystyle e^x = 0$ <-- Doesn't happen
or
$\displaystyle x - 1 = 0 \implies x = 1$
or
$\displaystyle x - 2 = 0 \implies x = 2$
So the critical points are x = 1 and x = 2.
-Dan
To find local extrema, it is when the derivative is equal to zero.
$\displaystyle f'(x)=\dfrac{e^x(1-x+x^2)-(-1+2x)(e^x)}{(1-x+x^2)^2}=\dfrac{e^x(x-1)(x-2)}{(1-x+x^2)^2}$
$\displaystyle 0=e^x(x-1)(x-2)$
therefore local max or min occur at x=1 and/or x=2
Now look at the interval <1;at 1;1-2;at 2;>2
you will see that at x=1 it is a local max and at x=2 it is a local min