[SimonHung]

How do I find the absolute maximum and minimum? Do I differentiate them and find f'(x) = 0 or do second derivative test?

[little_wizard123]

(Original post by SimonHung)


How do I find the absolute maximum and minimum? Do I differentiate them and find f'(x) = 0 or do second derivative test?

Well this would find all your local maxima and minima. As you're looking for the max or min on a specific range, it may not be a stationary point.

E.g. consider the absolute maxima and minima of x^2 on [0,3]. The minimum value on [0,3] could be found by setting dy/dx = 0, but the absolute maxima is x=3 which is not a stationary point.

[Hasufel]

if you derive the first, equate it to zero, and factor, you get:

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

this is defined where the NUMERATOR = 0, ie, x=1 (because -1 is not in the domain given), use this to get y...

[SimonHung]

(Original post by little_wizard123)
Well this would find all your local maxima and minima. As you're looking for the max or min on a specific range, it may not be a stationary point.

E.g. consider the absolute maxima and minima of x^2 on [0,3]. The minimum value on [0,3] could be found by setting dy/dx = 0, but the absolute maxima is x=3 which is not a stationary point.

So does this mean I have to draw the graph out? Are there any other approaches?

[little_wizard123]

(Original post by SimonHung)
So does this mean I have to draw the graph out? Are there any other approaches?

Approach it logically. You have a region [a,b]. You could have turning points anywhere on this region, including at the end points. Either a stationary point will be a minimum or maximum value on a region, or the end points will be. A max or min value on a,b) that is not a stationary point doesn't make sense as there would be a point 'next' to it both higher and lower.

[Boucly]

TL: DR version: find $x$ such that $f'(x)=0$. Find $f(x)$ for all those $x$. Find $f(a)$ and $f(b)$ where $a$ and $b$ are end points. (0 and 2 in the first question.) Now find the greatest one and the smallest one out of those $f(x)$ and $f(a)$ and $f(b)$.

Most things are said already. This is just to elaborate a bit.

Differentiation does not work (is not formally defined) on end points
Think of differentiation as drawing the tangent line at a point -- a "nice" line "balanced" on the graph at that point. Now, those bracketed domains, say $[a,b]$, mean that the points before $a$ and after $b$ are removed, so you can't "balance" a nice unique line on the end points.
Now to make this obvious, let $f(x)=x^2$ (or any nice function you want), but I make the domain $[1,1]$. Yes, a single point. Now I want you to draw me the tangent on that single point. How? You can't. It's an end point.

What does this have to do with the question? Now, the general way to tackle this question is to find points that just *may* be the biggest or the smallest and test them. To test them, you use $f$ on them, and find the biggest and the smallest out of those.
Okay, I'll just... Wait, wait. I want to type more. Another way to look at this is to remove all the points that cannot possibly be the answer and test the remaining ones. This is better since you cannot be sure you have found *all* possible ones.
You know that a point cannot possibly be a global maximum if it is not stationary. So, you can actually say that any points whose derivative that are not zero is not an answer (as you have rightly assumed, probably). But of couse, end points do not have a derivative defined. So the list of possible answers remains as: stationary points+end points, as noted by others already.

(Original post by SimonHung)
Do I differentiate them and find f'(x) = 0 or do second derivative test?

You will need to find values of $x$ such that $x$ is indeed zero. For the second derivative test, suppose you found it and know it is a local maximum (or minimum), what will you do then? You need to test if it is bigger than other maximum. Might as well skip the second derivative and compare its value of $f$ with the other maximums'.

[bijesh12]

First you find critical points of the function, which is a point (c), such that f '(c) does not exist or f '(c)=0 So we find all these values, where f '(c)=0 or where it don't exist. We then substitute these critical values into the original function, aswell as the endpoints. Whichever is the highest value, is the absolute maximum and the lowest value is the absolute minimum. First example :

f(x) = x / (x^2 + 1) in the interval [0,2]

firstly f(0) = 0 / (0 +1)
=0

f(2) = 2 / (4 + 1)
=2/5

Now find critical values

f'(x) =

Where does f'(x) not exist ? f '(x) exists for all real values
where does f'(x)= 0, is when x=1

Critical Points (at x=1)

so f(1)= 1/2

The absolute maximum,is the output of f(x) is greatest out of the endpoints and critical values, which is x=1, so 1/2.

The absolute minimum, is the output of f(x) is lowest out of the endpoints and critical values, which is when x=0.

Absolute Max = (1, 1/2)
Absolute Min = (0,0)

[SimonHung]

Guys, for part (b) I got:


Do I just treat it as 1-(1/4)x^2 =0 since e^-x^2/8 can be placed at the denominator?

[Hasufel]

this can also be factorized as: [(x+2)(x-2)]/ [-4Exp(x^2/8)] which has critical points at x=-2, x=2

left end point gives a negative -1/Exp[1/8]

right end point: 4/e^2

what is y at x=2? (-2 not in domain)

EDIT: largest/smallest of these 3 is the absolute min/max over the given domain