I need to find absolute max and min...please let me know how.

1.) f(x) = x + sin x on [0, π]      take the derivative and set to 0

f' '(x)  =  1 + cosx  = 0      subtract 1 from both sides

cosx  = -1       and this happens at the end point of the interval =  π

Using the second derivative as a critical point will not help us determine if this is a max or min because ..... -sin (π)  = 0

So...... testing two points on either side of π  and putting them into the first derivative will tell us if we might have a min or max....I'll choose 3π/4   and 5π/4

1 + cos(3π/4)  is positive

And

1 + cos(5π/4) is positive

Thus......this graph has no max or min on the requested interval......instead, π  is an inflection point where the graph changes from  concave down to concave up  [ this would be verified by putting the two chosen points into the second derivative ]

Here's a graph, Yura_chan that shows this :https://www.desmos.com/calculator/emx6sprsis

2.) f(x)= x^2/(x-1) on [-1, 1/2]     take the second derivative, set it to 0

f '(x)  =  [ 2x (x -1) - x^2 * 1] / (x -1))^2  = 0       multiply through by (x -1)^2, simplify the numerator

2x^2 - x^2 - 2x = 0

x^2 - 2x  = 0     factor

x(x -2)  = 0      setting each factor to 0, we have two critical points x = 0 , and x =2....we don't care about the second value since it's outside the interval of interest

We could take the second derivative, but it's kinda' messy...instead let's just choose 2 points around 0 and put them in the first derivative to see what the slope is at each value...I'll choose -1  and 1/2

Here's the first derivative again

f' (x)  =  [x^2 - 2x] / (x -1)^2

Evalutating this at x =   -1 we get  3/4

Evaluating this at  x = 1/2, we get -3

This tells us that we have a maximum at 0

Yura_chan, here's the graph that cofirms this : https://www.desmos.com/calculator/tv2ainbnca

BTW.....this graph is not continuous at x = 1......the other critical point of x = 2 produces a minimum