Extrema: Maxima & Minima
Last updated
Last updated
Extrema are one type of Critical points, which includes Maxima & Minima. And there're two types of Max and Min, Global Max & Local Max, Global Min & Local Min. We can all them Global Extrema or Local Extrema.
And actually we can call them in different ways, e.g.:
Global Max & Local Max or in short of glo max & loc max
Absolute Max & Relative Max or in short ofabs max & rel max
We need two kind of conditions to identify the Max or Min. Now If we have a Non-Endpoint Minimum or Maximum point at a, then it must satisfies these conditions:
Geometric condition: (It should be understood in a more intuitive way)
in the interval [a-h, a+h] there's no point above or below f(a) or
f'(a-h) & f'(a+h) have different sign, one negative another positive.
Derivative condition:
f'(a) = 0 or
f'(a) is undefined
Refer to Khan academy lecture: Finding critical points
We just need to assume f'(x) = 0 or f'(x) is undefined, and solve the equation to see what x value makes it then.
Increasing & Decreasing IntervalsWe can easily tell at a point of a function, it's at the trending of increasing or decreasing, by just looking at the instantaneous slope of the point, aka. the derivate. If the derivative, the slope is positive, then it's increasing. Otherwise it's decreasing.
Just been said above, we assume at point a, it's value is f(a). So the slope of it is f'(a). And if f'(a) < 0, then it's decreasing; If f'(a) > 0, then it's increasing.
Finding a decreasing or increasing intervalIt's just doing the same thing in the opposite way. For find a decreasing interval, we assume f'(x) < 0, and by solving the inequality equation we will get the interval.
Strategy:
Get Critical points.
Separate intervals according to critical points, undefined points and endpoints.
Try easy numbers in EACH intervals, to decide its TRENDING (going up/down):
If f'(x) > 0 then the trending of this interval is Increasing.
If f'(x) < 0 then the trending of this interval is Decreasing.
Set f'(x) = 0 or undefined, get x = -2 or -1/3
Separate intervals to (-∞, -2, -1/3, ∞)
Try some easy numbers in each interval: -3, -1, 0:
How to find Relative ExtremaRemember that an
Absolute extremeis also aRelative extreme.
Refer to khan: Worked example: finding relative extrema Refer to Khan Academy article: Finding relative extrema
Strategy:
Get Critical points.
Separate intervals according to critical points, undefined points and endpoints.
Try easy numbers in EACH intervals, to decide its TRENDING (going up/down).
Decide each critical point is Max, Min or Not Extreme.
Refer to an awesome article: Using calculus to learn more about the shapes of functions
Set f'(x)=0 or undefined, get x=0 or -2 or 1
Separate intervals to (-∞, -2, 0, 1, ∞)
Try easy number in each interval: -3, -1, 0.5, 2 and get the trendings:
Identify critical points' concavity:
How to find Absolute ExtremaRefer to Khan academy article: Absolute minima & maxima review Refer to Khan academy lecture: Finding absolute extrema on a closed interval
Strategy:
Find all Relative extrema
Get Critical points.
Separate intervals according to critical points & endpoints.
Try easy numbers in EACH intervals, to decide its TRENDING (going up/down).
Decide each critical point is Max, Min or Not Extreme.
Input all the extreme point into original function f(x) and get extreme value.
Set g'(x) = 0 or undefined get x=0
Separate intervals to [-2, 0, 3] according to the critical point & endpoints of the given condition.
Try some easy numbers of each interval -1, 2 into g'(x)
Get the trending of each interval: (+, +)
So the minimum must be the left endpoint of the given condition, which is x=-2.
Differentiate to set f'(x)=0 and got x=0, -1, 1
Separate intervals to (-∞, -1, 0, 1, +∞)
Try some easy numbers in each interval for f'(x) and got the signs: -, +, -, +
According to the signs we know that -, + gives us a Relative maximum
But at the end it's + again, and the interval is going up to +∞, means f(x) will go infinitely high.
So there's NO Absolute maximum.
Solve:
Solve:
Solve:
Solve:
