Math - Absolute max/min?

Question

Does anyone know anything about absolute max/min word problems? examples, solutions, & such

Answer 1

Absolute min/max problems usually deal with a function on a closed interval. If you're asked to find the minimum and maximum of f(x) on a closed interval [a,b], you would test the endpoints f(a) and f(b), find the derivative and make it equal to 0, and then test that point as well.

The highest value out of f(a), f(b), and f([any results after taking the derivative]) will be the absolute max, and the lowest, the absolute min.

Example: f(x) = 3x^2 + 2x + 5, on the interval [-1,1]
Find the absolute min and absolute max.

First step: calculate f(-1) and f(1)
f(-1) = 3(-1)^2 + 2(-1) + 5 = 3 - 2 + 5 = 6
f(1) = 3(1)^2 + 2(1) + 5 = 3 + 2 + 5 = 10

Now, calculate f'(x)
f'(x) = 6x + 2
Now, make f'(x) 0.
0 = 6x + 2
-2 = 6x
-1/3 = x

Notice that x = -1/3 lies in the interval [-1,1]. Now, we must calculate f(-1/3).

f(x) = 3x^2 + 2x + 5
f(-1/3) = 3*(-1/3)^2 + 2(-1/3) + 5 = 3(1/9) - 2/3 + 5
= 1/3 - 2/3 + 5 = -1/3 + 5 = 14/3

Therefore, we have
f(-1) = 6
f(1) = 10
f(-1/3) = 14/3

The biggest is clearly f(1), and the absolute maximum is (1,10). The smallest is f(-1/3), and the absolute min is (-1/3 , 14/3).

Answer 2

The idea is that a continuous function that increases and then decreases has a local maximum in the middle, and a function that decreases and then increases has a minimum between.

The way to determine is at the point where the hill or valley exists the derivative of that function will go to zero ... the middle between a positive rate of change and a negative rate of change or vice-versa.

The way you find these points if to differentiate the function and find the zeros for it ... there may be more than one.

Answer 3

a million.) you identify the spinoff f ' (x) and the 2d spinoff f '' (x). to get the 2d spinoff you're taking the first spinoff and derive it once back. now you've the traditional funktion f(x) it is given, its spinoff and its 2d spinoff. 2.) you equalise the spinoff with 0 . ( meaning f ' (x) = 0 ) then you actually remedy the x of this equation. all recommendations for x are potential minimum or optimal. 3.) now you opt to envision no matter if it really is a minimum or a optimal or none of both and to finish that that you want the 2d spinoff. you placed your recommendations for x in the 2d spinoff and examine no matter if it really is larger than 0 or smaller than 0. no matter if it really is larger you've a minimum. no matter if it really is smaller you've a optimal. no matter if it really is 0 you could't be sure no matter if it really is or no longer yet it is very uncommon to ensue. ultimately you placed the x fee into the traditional funktion to get the y fee e.g. a million.) f(x) = x^2 (favourite funktion ) f ' (x) = 2x (spinoff) f '' (x) = 2 ( 2d spinoff) 2.) f ' (x) = 0 --> 2x = 0 --> x = 0 0 is the actually answer for that equation and potential minimum or optimal. 3.) puttin 0 in the 2d spinoff f '' (0) = 2 meaning it really is a minimum. putting it in the funktion: f(0)= 0^2 =0 the minimum of this funktion is at (0/0)