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(1)The primary way to find extrema (maxima or minima) of functions is to use calculus. All extrema occur at critical values of the function -- those values where the first derivative is zero or fails to exist.
You find the first derivative, check each location where the derivative fails to exist. Then you set the first derivative equal to zero and solve for x. Once you have x, you evaluate the function at x to obtain the value.
You can use the first derivative test to check whether you have a local maximum, local minimum, or neither (occasionally an inflection point)
(2) You can locate extrema of some functions without calculus. For example, quadratics with a positive leading coefficient will have an absolute minimum, while quadratics with a negative leading coefficient have an absolute maximum. These extreme are located at the vertex of the parabola. The x coordinate of the extrema will be at `(-b)/(2a)` when the function is written in standard form `ax^2+bx+c=0`
(3) You could graph the function. Without calculus, or special cases like parabolas, you won't be able to prove you have an extremum, but you can get a reasonable approximation. With many graphing utilities you can locate an extremum on a stated interval.
This question is rather broad -- I hope this helped.
A very easy way of finding the minimum or maximum value of a function is by the y-value of the vertex.
e.g if the vetex is (x,y), the maximum or minimum value is y.
if a>o, the parabola opens up, so its minimum(y>=the y value of the vertex)
if a<0, the parabola opens down, so, its maximum.(y<= the y value of the vertex)
The final answer is written as: the maximum or minimum value is.....(the y value of the vertex) when x is......(the x value of the vertex).
I hope this helps you.
Sometimes one of the equations used to determine the function has a limitation which assists in finding the roots (or x-intercepts).
For example: If you have 50 m of wire fence and you need to maximize the area. You would use the area equation (A = width x length), and the perimeter equation (2xwidth + 2xlength = 50) to assist you with the determination of the function.
Letting width = W and length = L, you would need to:
1. solve for one of the variables
e.g. 2W + 2L = 50
W = (50 - 2L)/2 = 25 - L
2. substitute into the other equation
e.g. A = WL
= 25L - L^2 (this is your function, note it is concave down since a < 0, so we confirm that we will determine the maximum value of Area for given value of L at the vertex)
Now, to the part where you observe the constraint, or limitation. You know that width can not be zero nor negative so therefore length must satisfy 0 < L < 25. This provides you automatically with the axis of symmetry of L = 25/2 = 12.5.
You can now substitute L = 12.5 into the area function to calculate the maximum of A = 156 m^2.
In summary, look at the equations and try to observe any limitations on the variables involved.
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