Write a paragraph describing your durga puja puja holidays in such a manner that all the rules of the Subject- Verb Agreement must be applied while writing the paragraph.
Note:- The sentences woven in the paragraph must follow one or the other rule of the Subject- Verb Agreement. Care has to be taken that all the rules must be applied in the paragraph.
Answers
Answer:Durga Pooja is a Hindu festival celebration of the Mother Goddess and the victory of the warrior Goddess Durga over the demon Mahisasura. The festival represents female power as ‘Shakti’ in the Universe. It is a festival of Good over Evil. Durga Pooja is one of the greatest festivals of India. In addition to being a festival for the Hindus, it is also time for a reunion of family and friends, and a ceremony of cultural values and customs.
Explanation: The significance of Durga Pooja
While the ceremonies bring observance of fast and devotion for ten days, the last four days of the festival namely Saptami, Ashtami, Navami, and Vijaya-Dashami are celebrated with much sparkle and magnificence in India, especially in Bengal and overseas.
The Durga Pooja celebrations differ based on the place, customs, and beliefs. Things differ to the extent that somewhere the festival is on for five days, somewhere it is for seven and somewhere it is for complete ten days. Joviality begins with ‘Shashti’ – sixth day and ends on the ‘VijayaDashmi’ – the tenth day.
Answer:
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10
MAXIMUM AND MINIMUM
VALUES
The turning points of a graph
The turning points
WE SAY THAT A FUNCTION f(x) has a relative maximum value at x = a,
if f(a) is greater than any value immediately preceding or follwing.
We call it a "relative" maximum because other values of the function may in fact be greater.
We say that a function f(x) has a relative minimum value at x = b,
if f(b) is less than any value immediately preceding or follwing.
Again, other values of the function may in fact be less. With that understanding, then, we will drop the term relative.
The value of the function, the value of y, at either a maximum or a minimum is called an extreme value.
Now, what characterizes the graph at an extreme value?
The slope is 0
The tangent to the curve is horizontal. We see this at the points A and B. The slope of each tangent line -- the derivative when evaluated at a or b -- is 0.
f '(x) = 0.
Moreover, at points immediately to the left of a maximum -- at a point C -- the slope of the tangent is positive: f '(x) > 0. While at points immediately to the right -- at a point D -- the slope is negative: f '(x) < 0.
In other words, at a maximum, f '(x) changes sign from + to − .
At a minimum, f '(x) changes sign from − to + . We can see that at the points E and F.
We can also observe that at a maximum, at A, the graph is concave downward. (Topic 14 of Precalculus.) While at a minimum, at B, it is concave upward.
A value of x at which the function has either a maximum or a minimum is called a critical value. In the figure --
Critical values
-- the critical values are x = a and x = b.
The critical values determine turning points, at which the tangent is parallel to the x-axis. The critical values -- if any -- will be the solutions to f '(x) = 0.
Example 1. Let f(x) = x2 − 6x + 5.
Are there any critical values -- any turning points? If so, do they determine a maximum or a minimum? And what are the coördinates on the graph of that maximum or minimum?
Solution. f '(x) = 2x − 6 = 0 implies x = 3. (Lesson 9 of Algebra.)
x = 3 is the only critical value. It is the x-coördinate of the turning point. To determine the y-coördinate, evaluate f at that critical value -- evaluate f(3):
f(x) = x2 − 6x + 5
f(3) = 32 − 6· 3 + 5
= −4.
The extreme value is −4. To see whether it is a maximum or a minimum, in this case we can simply look at the graph.
Critical values
f(x) is a parabola, and we can see that the turning point is a minimum.
By finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (3, −4).
But we will not always be able to look at the graph. The algebraic condition for a minimum is that f '(x) changes sign from − to + . We see this at the points E, B, F above. The value of the slope is increasing.
Now to say that the slope is increasing, is to say that, at a critical value, the second derivative (Lesson 9) -- which is rate of change of the slope -- is positive.
Again, here is f(x):
f(x) = x2 − 6x + 5.
f '(x) = 2x − 6.
f ''(x) = 2.
f '' evaluated at the critical value 3 -- f''(3) = 2 -- is positive. This tells us algebraically that the critical value 3 determines a minimum.
Sufficient conditions
Extreme values
We can now state these sufficient conditions for extreme values of a function at a critical value a:
The function has a minimum value at x = a if f '(a) = 0
and f ''(a) = a positive number.
The function has a maximum value at x = a if f '(a) = 0
and f ''(a) = a negative number.
In the case of the maximum, the slope of the tangent is decreasing -- it is going from positive to negative. We can see that at the points C, A, D.
Example 2. Let f(x) = 2x3− 9x2 + 12x − 3.
Are there any extreme values? First, are there any critical values -- solutions to f '(x) = 0 -- and do they determine a maximum or a minimum? And what are the coördinates on the graph of that maximum or minimum? Where are the turning points?
Solution. f '(x) = 6x2 − 18x + 12 = 6(x2 − 3x + 2)
= 6(x − 1)(x − 2)
= 0
implies:
x = 1 or x = 2.
(Lesson 37 of Algebra.)
Those are the critical values. Does each one determine a maximum or does it determine a minimum? To answer, we must evaluate the second derivative at each value.
f '(x) = 6x2 − 18x + 12.
f ''(x) = 12x − 18.
f ''(1) = 12 − 18 = −6.
The second derivative is negative. The function therefore has a maximum at x = 1.
To find the y-coördinate -- the extreme value -- at that maximum we evaluate f(1):
f(x) = 2x3− 9x2 + 12x − 3
f(1) = 2 − 9 + 12 − 3
= 2.
The maximum occurs at the point (1, 2).
Next, does x = 2 determine a maximum or a minimum?
f ''(x) = 12x − 18.
f ''(2) = 24 − 18 = 6.
The second derivative is positive. The function therefore has a minimum at x = 2.
To find the y-coördinate -- the extreme value -- at that minimum, we evaluate f(2):
f(x) = 2x3 − 9x2 + 12x − 3.
f(2) = 16 − 36 + 24 − 3
= 1.
The minimum occurs at the point (2, 1).