Maxima and Minima
Maxima and Minima - Let f(x) be a real valued function defined on an interval I.Then, f(x) is said to have the maximum value in the interval I,if there exists a point a in I such that f(x) $\leq$ f(a) for all x `in` I.The number f(a) is called the maxima or the maximum value of f(x) in the interval I and the point a is called a point of maxima of f in the interval I.
Let f(x) be a real valued function defined on an interval I.Then, f(x) is said to have the maximum value in the interval I,if there exists a point a in I such that f(x) $\leq$ f(a) for all x `in` I.The number f(a) is called the maxima or the maximum value of f(x) in the interval I and the point a is called a point of maxima of f in the interval I.
Let f(x) be a real valued function defined on an interval I.Then f(x) is said to have the minimum value in interval I, if there exists a point a `in` I such that f(x) $\geq$ f(a) for all x`in`I.The number f(a) is called the minima or minimum value of f(x) in the interval I and the point a is called a point of minima of f in the interval I.
Example : Find the maxima of a function f given by f(x) = -(x -1)2 + 10
Solution : Clearly, domain of f = `RR`= (-`oo` , `oo` ).
Here, -(x-1)2 $\leq$ 0 for all x `in` `RR` `rArr` -(x-1)2 + 10 $\leq$ 10 for all x `in` `RR`
`rArr` f(x) $\leq$ 10 for all x `in` `RR` [ f(1) = -(1-1)2 + 10 = 10 ]
`rArr` f(x) $\leq$ f(1) for all x `in` `RR`
Hence f(1) = 10 is the maxima of function f and the point of maxima of f is x = 1.The graph of function f is shown below:
Example:Find the minima of a function f given by f(x) = x2 + 5
Solution: Clearly, domain f = `RR` = (-`oo` , `oo` )
Here, x2 $\geq$ 0 for all x `in` `RR` `rArr` x2 + 5 $\geq$ 5 for all x `in` `RR`
`rArr` f(x) $\geq$ 5 for all x `in` `RR` [ f(0) = 0 + 5 = 5 ]
`rArr` f(x) $\geq$ f(0) for all x `in` `RR`
Hence minima of the function f(x) is 5 and the point of minima of f is x = 0.The graph is shown below:
In the above discussion, we find that function f(x) = -(x - 1)2 + 10, has a maximum value but it does not attain the minimum value. Here f(x) is a decreasing function.Similarly, the function f(x) = x2 + 5 attains a minimum value 5 at x = 0,but it does not attain the maximum value at any point in its domain. From the graph, we find that f(x) is an increasing function.
Let us now consider the function f(x) = sin x defined in the interval [0,2`pi`] clearly,
-1 $\leq$ sin x $\leq$ 1 for all x`in` [0, 2`pi`]
Also, f($\frac{\pi}{2}$) = 1 and f($\frac{3\pi}{2}$) = -1
Therefore, f($\frac{3\pi}{2}$) $\leq$ f(x) $\leq$ f($\frac{\pi}{2}$) for all x`in` [0,2`pi`]
Thus,f(x) attains both the maximum value 1 and the minimum value -1 in the interval [0,2`pi`] points x = $\frac{\pi}{2}$ and x = $\frac{3\pi}{2}$ are respectively the points,maximum and minimum values of f in the interval [0,2`pi`] this is also evident from the graph shown below:
Maxima and Minima - Let f(x) be a real valued function defined on an interval I.Then, f(x) is said to have the maximum value in the interval I,if there exists a point a in I such that f(x) $\leq$ f(a) for all x `in` I.The number f(a) is called the maxima or the maximum value of f(x) in the interval I and the point a is called a point of maxima of f in the interval I.
Let f(x) be a real valued function defined on an interval I.Then, f(x) is said to have the maximum value in the interval I,if there exists a point a in I such that f(x) $\leq$ f(a) for all x `in` I.The number f(a) is called the maxima or the maximum value of f(x) in the interval I and the point a is called a point of maxima of f in the interval I.
Let f(x) be a real valued function defined on an interval I.Then f(x) is said to have the minimum value in interval I, if there exists a point a `in` I such that f(x) $\geq$ f(a) for all x`in`I.The number f(a) is called the minima or minimum value of f(x) in the interval I and the point a is called a point of minima of f in the interval I.
Example : Find the maxima of a function f given by f(x) = -(x -1)2 + 10
Solution : Clearly, domain of f = `RR`= (-`oo` , `oo` ).
Here, -(x-1)2 $\leq$ 0 for all x `in` `RR` `rArr` -(x-1)2 + 10 $\leq$ 10 for all x `in` `RR`
`rArr` f(x) $\leq$ 10 for all x `in` `RR` [ f(1) = -(1-1)2 + 10 = 10 ]
`rArr` f(x) $\leq$ f(1) for all x `in` `RR`
Hence f(1) = 10 is the maxima of function f and the point of maxima of f is x = 1.The graph of function f is shown below:
Example:Find the minima of a function f given by f(x) = x2 + 5
Solution: Clearly, domain f = `RR` = (-`oo` , `oo` )
Here, x2 $\geq$ 0 for all x `in` `RR` `rArr` x2 + 5 $\geq$ 5 for all x `in` `RR`
`rArr` f(x) $\geq$ 5 for all x `in` `RR` [ f(0) = 0 + 5 = 5 ]
`rArr` f(x) $\geq$ f(0) for all x `in` `RR`
Hence minima of the function f(x) is 5 and the point of minima of f is x = 0.The graph is shown below:
In the above discussion, we find that function f(x) = -(x - 1)2 + 10, has a maximum value but it does not attain the minimum value. Here f(x) is a decreasing function.Similarly, the function f(x) = x2 + 5 attains a minimum value 5 at x = 0,but it does not attain the maximum value at any point in its domain. From the graph, we find that f(x) is an increasing function.
Let us now consider the function f(x) = sin x defined in the interval [0,2`pi`] clearly,
-1 $\leq$ sin x $\leq$ 1 for all x`in` [0, 2`pi`]
Also, f($\frac{\pi}{2}$) = 1 and f($\frac{3\pi}{2}$) = -1
Therefore, f($\frac{3\pi}{2}$) $\leq$ f(x) $\leq$ f($\frac{\pi}{2}$) for all x`in` [0,2`pi`]
Thus,f(x) attains both the maximum value 1 and the minimum value -1 in the interval [0,2`pi`] points x = $\frac{\pi}{2}$ and x = $\frac{3\pi}{2}$ are respectively the points,maximum and minimum values of f in the interval [0,2`pi`] this is also evident from the graph shown below:
Let f(x) be a real valued function defined on an interval I.Then, f(x) is said to have the maximum value in the interval I,if there exists a point a in I such that f(x) $\leq$ f(a) for all x `in` I.The number f(a) is called the maxima or the maximum value of f(x) in the interval I and the point a is called a point of maxima of f in the interval I.
Let f(x) be a real valued function defined on an interval I.Then f(x) is said to have the minimum value in interval I, if there exists a point a `in` I such that f(x) $\geq$ f(a) for all x`in`I.The number f(a) is called the minima or minimum value of f(x) in the interval I and the point a is called a point of minima of f in the interval I.
Example : Find the maxima of a function f given by f(x) = -(x -1)2 + 10
Solution : Clearly, domain of f = `RR`= (-`oo` , `oo` ).
Here, -(x-1)2 $\leq$ 0 for all x `in` `RR` `rArr` -(x-1)2 + 10 $\leq$ 10 for all x `in` `RR`
`rArr` f(x) $\leq$ 10 for all x `in` `RR` [ f(1) = -(1-1)2 + 10 = 10 ]
`rArr` f(x) $\leq$ f(1) for all x `in` `RR`
Hence f(1) = 10 is the maxima of function f and the point of maxima of f is x = 1.The graph of function f is shown below:
Example:Find the minima of a function f given by f(x) = x2 + 5
Solution: Clearly, domain f = `RR` = (-`oo` , `oo` )
Here, x2 $\geq$ 0 for all x `in` `RR` `rArr` x2 + 5 $\geq$ 5 for all x `in` `RR`
`rArr` f(x) $\geq$ 5 for all x `in` `RR` [ f(0) = 0 + 5 = 5 ]
`rArr` f(x) $\geq$ f(0) for all x `in` `RR`
Hence minima of the function f(x) is 5 and the point of minima of f is x = 0.The graph is shown below:
In the above discussion, we find that function f(x) = -(x - 1)2 + 10, has a maximum value but it does not attain the minimum value. Here f(x) is a decreasing function.Similarly, the function f(x) = x2 + 5 attains a minimum value 5 at x = 0,but it does not attain the maximum value at any point in its domain. From the graph, we find that f(x) is an increasing function.
Let us now consider the function f(x) = sin x defined in the interval [0,2`pi`] clearly,
-1 $\leq$ sin x $\leq$ 1 for all x`in` [0, 2`pi`]
Also, f($\frac{\pi}{2}$) = 1 and f($\frac{3\pi}{2}$) = -1
Therefore, f($\frac{3\pi}{2}$) $\leq$ f(x) $\leq$ f($\frac{\pi}{2}$) for all x`in` [0,2`pi`]
Thus,f(x) attains both the maximum value 1 and the minimum value -1 in the interval [0,2`pi`] points x = $\frac{\pi}{2}$ and x = $\frac{3\pi}{2}$ are respectively the points,maximum and minimum values of f in the interval [0,2`pi`] this is also evident from the graph shown below:
Let f(x) be a real valued function defined on an interval I.Then, f(x) is said to have the maximum value in the interval I,if there exists a point a in I such that f(x) $\leq$ f(a) for all x `in` I.The number f(a) is called the maxima or the maximum value of f(x) in the interval I and the point a is called a point of maxima of f in the interval I.
Let f(x) be a real valued function defined on an interval I.Then f(x) is said to have the minimum value in interval I, if there exists a point a `in` I such that f(x) $\geq$ f(a) for all x`in`I.The number f(a) is called the minima or minimum value of f(x) in the interval I and the point a is called a point of minima of f in the interval I.
Example : Find the maxima of a function f given by f(x) = -(x -1)2 + 10
Solution : Clearly, domain of f = `RR`= (-`oo` , `oo` ).
Here, -(x-1)2 $\leq$ 0 for all x `in` `RR` `rArr` -(x-1)2 + 10 $\leq$ 10 for all x `in` `RR`
`rArr` f(x) $\leq$ 10 for all x `in` `RR` [ f(1) = -(1-1)2 + 10 = 10 ]
`rArr` f(x) $\leq$ f(1) for all x `in` `RR`
Hence f(1) = 10 is the maxima of function f and the point of maxima of f is x = 1.The graph of function f is shown below:
Example:Find the minima of a function f given by f(x) = x2 + 5
Solution: Clearly, domain f = `RR` = (-`oo` , `oo` )
Here, x2 $\geq$ 0 for all x `in` `RR` `rArr` x2 + 5 $\geq$ 5 for all x `in` `RR`
`rArr` f(x) $\geq$ 5 for all x `in` `RR` [ f(0) = 0 + 5 = 5 ]
`rArr` f(x) $\geq$ f(0) for all x `in` `RR`
Hence minima of the function f(x) is 5 and the point of minima of f is x = 0.The graph is shown below:
In the above discussion, we find that function f(x) = -(x - 1)2 + 10, has a maximum value but it does not attain the minimum value. Here f(x) is a decreasing function.Similarly, the function f(x) = x2 + 5 attains a minimum value 5 at x = 0,but it does not attain the maximum value at any point in its domain. From the graph, we find that f(x) is an increasing function.
Let us now consider the function f(x) = sin x defined in the interval [0,2`pi`] clearly,
-1 $\leq$ sin x $\leq$ 1 for all x`in` [0, 2`pi`]
Also, f($\frac{\pi}{2}$) = 1 and f($\frac{3\pi}{2}$) = -1
Therefore, f($\frac{3\pi}{2}$) $\leq$ f(x) $\leq$ f($\frac{\pi}{2}$) for all x`in` [0,2`pi`]
Thus,f(x) attains both the maximum value 1 and the minimum value -1 in the interval [0,2`pi`] points x = $\frac{\pi}{2}$ and x = $\frac{3\pi}{2}$ are respectively the points,maximum and minimum values of f in the interval [0,2`pi`] this is also evident from the graph shown below:
By a local maximum(or local minimum) value of a function at a point x = a we mean the greatest (or the least) value in the neighborhood of point x = a and not the maximum (or the minimum) in the domain of the function.
In fact a function may have any number of points of local maximum(or local minimum) and even a local minimum value may be greater than a local maximum value.
$\Rightarrow$ If f is a differentiable function defined on an interval I and a `in` I.Then,
- x = a is a point of local maximum value of f, if
- f1(a) = 0 and,
- f1(x) changes sign from positive to negative as x increases through a,i.e. f1(x) > 0 at every point sufficiently close to and to the left of a,and f1(x) < 0 at every point sufficiently close to and to the right of a.
- x = a is a point of local minimum value of f, if
- f1(a) = 0 and,
- f1(x) changes sign from negative to positive as x increases through a,i.e. f1(x) < 0 at every point sufficiently close to and to the left of a, and f1(x) > 0 at every point sufficiently close to and to the right of a.
Steps to find local maxima or local minima of differentiable functions:
- Put y = f(x)
- Find `dy/dx`
- Put `dy/dx` = 0 and solve this equation for x.Let c1,c2,c3,...,cn be the roots of this equation.These are the possible points where the function can attain a local maximum or a local minimum.So we test the function at each of these points.
- Consider x = c1 .
- If `dy/dx` changes its sign from positive to negative as x increases through c1 , then the function attains a local maximum at x = c1
- If `dy/dx` changes its sign from negative to positive as x increases through c1 , then the function attains a local minimum at x = c1
- If `dy/dx` does not change its sign as x increases through c1 , then x = c1 is neither a point of local maxima nor a point of local minima.x = c1 is a point of inflexion.
- If f(x) is continuous function in its domain,then at least one maxima and minima must lie between two equal values of x.
- Maxima and minima occur alternately, that is, between two maxima there is one minimum and vice verse.
- If f(x)`->` `oo` as x`->` a or b and f1(x) = 0 only for one value of x (say c) between a and b,then f(c) is necessarily the minimum and the least value. If f(x)`->` -`oo` as x`->` a or b, then f(c) is necessarily the maximum and the greatest value.
Maxima and minima in a closed interval:
Let y = f(x) be a function defined on [a,b]. By local maxima and minima of a function at a point c `in` [a,b] we mean the greatest(or the least) value in the immediate neighborhood of x = c. It does not mean the greatest(or the least) of f(x) in the interval [a, b]. A function may have a number of local maxima or local minima in a given interval and even a local minimum may be greater than a relative maximum.
The following algorithm is the working rule for finding the absolute maximum and the absolute minimum of a function f defined on a closed interval [a, b].
Algorithm:
- Find f1(x)
- Put f1(x) = 0 and find the values of x. Let c1,c2,...,cn be the values of x.
- Take the maximum and minimum values out of the values f(a), f(c1),f(c2),...,f(cn), f(b).
The maximum and the minimum values obtained are respectively the largest or absolute maximum and the smallest or absolute minimum values of the function.
| More topics in Maxima and Minima | |
| Local Maxima and Minima | Absolute Maximum |
