Mathematics Modulus Function - Basics , Graph and Transformations

Please Wait... While Loading Full Video

Modulus Function - Basics , Graph and Transformations

Modulus Functions

Absolute value function (or modulus function)

`y= |x| = { tt((x, x ge 0),(-x, x <0))`

It is the numerical value of `x`

It is symmetric about `y`-axis"

Its domain is in `R` and range `[0, oo)`.

Properties of modulus functions

(i) `| x| le a => a le x le a ; (a ge 0)`

(ii) `|x| ge a => x le a` or `x ge a ; (a ge 0)`

(iii) `| |x| - | y | le |x ± y | le | x | + | y |`

Graph of | 2-3y| = | -2|x| +1 |

We willl start from the graph of y = x

After that we will draw the graph of y = 2 x

Then 3y = 2x

Then 3y = -2x

Then -3y = -2x

Then -3y + 2 = -2x

Then -3y +2 = -2x +1

Then -3y + 2 = -2 |x| + 1

Then -3y +2 = | -2 |x| +1 |

Then | -3y + 2 | = | -2 |x| +1 |

Q 2508545408

Plot `y = | x| + 2` and `y = | x| - 2`, with the help of `y =| x|`·

Solution:

`y =| x|` (modulus function) could be plotted as shown in Fig(1);

`=> y =| x| + 2` is shifted upwards by `2` units as shown in Fig (2)

also `y =| x|- 2` is shifted downwards by `2` units. as shown in Fig(3)

Q 2528645501

Plot `y =| x|, y = | x- 2|` and `y = | x + 2|`.

Solution:

As discussed `f(x) -> f(x - a)` ; shift towards right.

`=> y = | x - 2|` is shifted `'2'` units towards right. (Fig. 1)

`y = | x + 2|` is shifted `'2'` units towards left. (Fig. 2)

Q 2508645508

Sketch the curve `| x| + | y | = 1`

Solution:

As the graph for `y = 1- x` is ; (Fig. 1)

(i) `y = 1 - x -> y = 1 - | x |` (Fig. 2)

(ii) `y = 1- |x|-> |y| = 1 - |x|` . (Fig. 3)

Clearly above figure represents a square

Q 2508745608

Draw the graph for : `|y|=|2-1/(|x-1|)|`

Solution:

We know the graph for `y =1/x` is shown as; (Fig. 1)

(a) `y= 1/x -> y=1/(x-1)` (Fig.2)

(b) `y= 1/(x-1)-> y= 1/(|x-1|)` (Fig.3)

(c) `y=1/(|x-1|)-> y=-1/(|x-1|)` (Fig. 4)

(d) `y= -1/(|x-1|)-> y=2-1/(|x-1|)` (Fig. 5)

(e) `y=2-1/(|x-1|)-> y= |2-1/(|x-1|)|` (Fig. 6)

(f) `y=|2- 1/(|x-1|)| -> |y|= |2-1/(|x-1|)|` (Fig. 7)

Q 2518045800

Sketch the curve of the following : `|x|+|y| le 2`

Solution:

To sketch `| x | + | y | le 2`

Here; `|y|= 2-|x|` is plotted as;

From above graph of `|y| = 2 - | x |`, we can check shading of

`|y| le 2 -| x|` or `|x|+|y| le 2`.

as at `(0, 0) => 0 le 2` (true)

Q 2548045803

Sketch the curve; `| x + y | + | x -y | ge 4`

Solution:

`| x + y | + | x -y | ge 4 => |x| ge 2` and `|y| ge 2`

Thus, to shade the portion when `x le -2` or `x ge 2` and `yle -2` or `y ge 2`. Shown as in Fig.

Transformation of `y=x` to `y=2x`

Plot `y= sinx` and `y=sin(x/2)`

Here, `y=sin(x/2)` is to stretch (or expand) the graph of `sin x '2'` times along `x`-axis.

(i) Plotting `y=sin^(-1) x`

(ii) Plotting `y=sin^(-1) (x/3)`

`f(x)` transforms to `f(ax)`

i.e., `f(x)-> f(ax); a > 1`

Shrink (or contract) the graph of `f(x) 'a'` times along `x`-axis.

again `f(x)-> f(1/a x) ; a>1`

Stretch (or expand) the graph of `f(x) 'a'` times along `x`-axis.

Graphically it could be stated as shown in fig.

Transformation of `y=2x` to `3y=2x`

Plot `y=sin x` and `y= 1/2 sin x`

As we know;

`y=1/a f(x)`

`=>` shrink the graph of `f(x)``'a'` times along `y`-axis.

`:. y= 1/2 sin x`

`=>` shrink the graph of `f(x)'2'` times along `y`-axis.

Transformation of `3y=2x` to `3y=-2x`

`y=e^(-x)`

As `y=e^x` is known; then `y = e^(-x)` is the image in `y`-axis as plane mirror for `y = e^x` shown as ;

Transformation of `3y=-2x` to `-3y=-2x`

Plot the curve `y=-e^(x)`

As `y = e^x` is known ;

`:. y=-e^x` take image of `y = e^x` in the x-axis as plane mirror

Transformation of `-3y=-2x` to `-3y +2 =-2x`

Plot `y = | x| + 2` and `y = | x| - 2`, with the help of `y =| x|`·

`y =| x|` (modulus function) could be plotted as shown in Fig

`=> y =| x| + 2` is shifted upwards by `2` units as shown in Fig

also `y =| x|- 2` is shifted downwards by `2` units. as shown in Fig

Transformation of `-3y + 2 = -2x` to `-3y +2 =-2x + 1`

Plot `y =e^x + 1; y =e^x - 1`, with the help of `y =e^x`.

We know; `y =e^x` (exponential function) could be plotted as;

`y = e^x + 1`, is shifted upwards by `1` unit, shown as

Also `y = e^x - 1`, is shifted downwards by `1` unit, shown as

Transformation of `-3y + 2 = -2x` to `-3y +2 =-2 |x| + 1`

Transformation of `-3y + 2 = -2|x| + 1 ` to `-3y +2 = |-2 |x| + 1 |`

Sketch the graph for `y=|e^(-|x|)-1/2|`

As we know the graph for `y=e^(-x)` (Fig 10.2)

(i) `y=e^(-|x|)-> y=e^(-x) -1/2` (Fig 10.3)

(ii) `y=e^(-x)-1/2-> y=e^(-|x|) -1/2` (Fig 10.4)

(iii) `y=|e^(-|x|) -1/2|` (Fig 10.5)

`f(x)` transforms to `|f(x)| ;`

Here; `y = | f(x) |` is drawn in two steps.
(a) In the I step, leave the positive part of `f(x)` , {i.e., the part of `f(x)` above `x`-axis) as it is.
(b) In the II step, take the mirror image of negative part of f(x). {i.e., the part of f(x) below x-axis} in the x-axis as plane mirror.

or

Take the mirror image (in x-axis) of the portion of the graph of f(x) which lies below x-·axis.

or

Turn the portion of the graph of f(x) lying below x-axis by 180° about x-axis.

Transformation of `-3y + 2 = |-2|x| + 1| ` to `|-3y +2| = |-2 |x| + 1 |`