What are the absolute extrema of #f(x) =x/(x^2-x+1) in[0,3]#?

Question

What are the absolute extrema of #f(x) =x/(x^2-x+1)  in[0,3]#?

Paisley Johnson · Accepted Answer

Absolute minimum is #0# (at #x=0#) and absolute maximum is #1# (at #x=1#).

#f'(x) = ((1)(x^2-x+1)-(x)(2x-1))/(x^2-x+1)^2 = (1-x^2)/(x^2-x+1)^2# #f'(x)# is never undefined and is #0# at #x=-1# (which is not in #[0,3]#) and at #x=1#. Testing the endpoints of the intevral and the critical number in the interval, we find: #f(0) = 0# #f(1) = 1# #f(3) = 3/7# So, absolute minimum is #0# (at #x=0#) and absolute maximum is #1# (at #x=1#).

Samantha Adkison · Answer

undefined To find the absolute extrema of $ f(x) = \frac{x}{x^2 - x + 1} $ on the interval $[0, 3]$, you first need to check critical points and endpoints.

1. Find critical points by setting the derivative equal to zero and solving for $ x $.

$$ f'(x) = \frac{(x^2 - x + 1) - x(2x - 1)}{(x^2 - x + 1)^2} = 0 $$

Solve for $ x $ to find critical points.

2. Check endpoints of the interval $[0, 3]$, which are $ x = 0 $ and $ x = 3 $.

3. Evaluate $ f(x) $ at critical points and endpoints.

4. Determine the maximum and minimum values of $ f(x) $ from the values obtained in step 3.

These steps will help identify the absolute extrema of the function $ f(x) $ on the interval $[0, 3]$.