# `f(x) = x/(x^2 - x + 1), [0, 3]` Find the absolute maximum and minimum values of f on the given interval

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Given: `f(x)=x/(x^2-x+1), [0,3].`

Find the critical values by setting the first derivative equal to zero and solving for the x values. Find the derivative using the quotient rule.

`f'(x)=[(x^2-x+1)(1)-x(2x-1)]/(x^2-x+1)^2=0`

`x^2-x+1-2x^2+x=0`

`-x^2+1=0`

`x^2=1`

`x=+-sqrt(1)`

`x=+-1`

The critical values are x=1 and x=-1. Substitute the critical values and the endpoints of the interval [0, 3] in to the original f(x) function. Do NOT substitute in the x=-1 because it is not in the given interval [0, 3].

f(0)=0

f(1)=1

f(3)=3/7

Evaluate the f(x) values to determine the absolute maximum and absolute minimum.

The **absolute maximum** occurs at the coordinate **(1, 1)**.

The **absolute minimum** occurs at the coordinate **(0, 0)**.

`f(x)=x/(x^2-x+1)`

differentiating by applying quotient rule,

`f'(x)=((x^2-x+1)-x(2x-1))/(x^2-x+1)^2`

`f'(x)=(x^2-x+1-2x^2+x)/(x^2-x+1)^2`

`f'(x)=(-x^2+1)/(x^2-x+1)^2`

Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.

Now to find the critical numbers, solve for x for f'(x)=0.

`(-x^2+1)/(x^2-x+1)^2=0`

`-x^2+1=0`

`x^2=1`

`x=1 , -1`

Since -1 is not in the interval (0,3) , so we have to evaluate f(x) at 1,0 and 3.

`f(0)=0`

`f(3)=3/(3^2-3+1)=3/7`

`f(1)=1`

So the function has **absolute maximum=1 at x=1**

**Function has absolute minimum=0 at x=0**