Maximum and Minimum value of a function

Maximum and minimum value of a function can be determined by simplifying the function with given intervals of time.

Function is a concept which expresses the idea that one quantity completely determines another quantity. Here is one such problem for your practice and understand how the value of function varies with respect to given interval of time.

Question : Find the maximum and minimum value of the function y = x³ - x on the interval [-3,3]

On substituting the value of x as -3 and 3, maxima and minima of function are found.

Solution :

When x = -3

y = x³ - x

= (-3)³ - (-3)

= -27 + 3

= - 24

And when x = 3

y = (3)³ - 3

= 27 - 3

= 24

Hence maximum value of function is 24 and minimum is - 24

A simple problem on ratio's

Topic:- Ratio's

If we wish to divide a amount or a number based on specific parts we use ratios.This will help us to share the values to different people in different quantities.

Here is an example which explain this better.

Question:-

How to share 35 between two people with the ration of 1:6

Answer:-

The given ratio is 1:6

The sum of the ratios is 1+6=7

Let us assume it is divided between A and B

    1
A= ----- * 35
    7

  35
= -----
  7

=  5

    6
B= ----- * 35
    7

   210
= -------
    7

So A gets 5 and b gets 30

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Problem on Mixed fractions

Topic:-Mixed fractions

A fraction is a number that can represent part of a whole.
The earliest fractions were reciprocals of integers, symbols representing one half, one third, one quarter, and so on.A much later development were the common or vulgar fractions which are still used today, and which consist of a numerator and a denominator.

In this algebra help ,we are giving an example of mixed fraction problem.
Topic:- Mixed Fractions
Question:-

 
 
  3      10
8--- - 4---- = ? 
  5      15


Answer:-

  3      10
8--- - 4---- = ? 
  5      15

(5*8)+3      (15*4)+10
--------  -  ----------
  5              15

 43          70
------  -  ------- 
  5          15

Multiply the numarator and the 
denominator of the first fraction 
with 3 to make equal denomitors 


   43*3         70
 -------  -   ------
   5*3          15

 
    129       70
   -----  -  -----
    15        15

      129-70
    -----------
        15

          59        14
         ----  or 3------
          15        15

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Examples of Locus

Topic:- Locus

Locus refers to a collection of points.

It's generally used to define a figure

Example 1:


A circle is a locus of all points ,Which are equidistant from a fixed point ,called the center of the circle.


Example 2:

A line is the locus of all points equidistant from two fixed points or from two parallel lines.

Let us consider a line

C_____________________________D >

-   -   -   -   -   -  -
A_____________________________B >
 k  l  m  o  n  p  q

Here,AB is a collection of points k,l,m,n,o,p,q
So,AB is called as Locus of all these points.

Commom point:
All the points are equidistant from line CD.

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Using Shell or Disc Method to Find Volume of the Solid

Disc method and Shell(cylinder) method of integration are the two different methods of finding volume of solid of a revolution, using rectangular coordination system the functions are defined in terms of x in the below problem.

Topic : Disc or Cylinder Method of Finding Volume of the Sphere.

Problem : Use the disc or shell method to find the volume of the solid generated by revolving the regions bounded by the graphs of the equations about the x axis. y=x³ y=0, x=2

Solution :

y = x³ => ³√y = x
or (y)^1/3 = x
or x = y^1/3

Volume of a Solid by rotating about x-axis is given by:

V = 2π_a∫^bp(y)h(y) dy
here p(y)=y^1/3, h(y)=y
when x = 2 and y = 3³ = 8
So a = 0 and b = 8
Plugging in all the values in the formula, we get

V = 2π₀∫⁸(y)^1/3.y dy

= 2π₀∫⁸(y)^4/3 dy (as 1/3 + 1/1 = (1+3)/3 = 4/3)

= 2π[y^7/3/(7/3)₀]⁸ (as 4/3 + 1/1 = (4+3)/3 = 7/3)

= 2π[(8)^7/3/(7/3)- (0)^7/3/(7/3)]

= 2π[((2)³)^7/3/(7/3)]

= 2π(2⁷/(7/3))

= 2π(128/(7/3))

= 2 * 3 * 128 * π / 7

= 768π /7


So this how the volume of Solid of revolution is determined when the equations about the x axis.
For more help write to our calculus help.

Problem on Trigonometric Identity

Trigonometric Identities are the standard equations used to solve trigonometric problems.
Below is as one such problem.

Topic : Trigonometric Identity

A proof for Trigonometric Identity

Problem : Prove the given identity 2 Cos x - 2 Cos³ x = Sin x Sin 2x

Solution :

LHS : 2 Cos x - 2 Cos³ x

= 2 Cos x (1 - Cos²x)
= 2 Cos x . Sin² x
= 2 Cos x . Sin x . Sinx
= Sin x . 2 Sin x Cos x
= Sin x . Sin 2x (using Sin 2x = 2 Sinx Cox x)

So we proved 2 Cos x - 2 Cos³ x = Sin x Sin 2x

If you have any queries related to the proof please leave a comment and we will get back to you soon.

A Solved Problem on Simplification

Topic : Simplification

Problem : Simplify (c + 9)(c³ - 3c² - 7c)

Solution :
(c + 9)(c³ - 3c² - 7c)
= c^4 - 3c³ - 7c² + 9c³ - 27c² - 63c
= c^4 + 6c³ - 34c² - 63c