Mathematics Examples of Graph of `| x| ` and its Transformation :

Miscellaneous Example : 1 Draw the graph of ` |y|= |2-1/(|x-1|)|`

Step 1 : We know the graph for `y =1/x` is shown as

Step 2 : `y= 1/x -> y=1/(x-1)`

Step 3 : `y= 1/(x-1)-> y= 1/(|x-1|)`

Step 4 : `y=1/(|x-1|)-> y=-1/(|x-1|)`

Step 5 : `y= -1/(|x-1|)-> y=2-1/(|x-1|)`

Step 6 : `y=2-1/(|x-1|)-> y= |2-1/(|x-1|)|`

Step 7 : `y=|2- 1/(|x-1|)| -> |y|= |2-1/(|x-1|)|`

Miscellaneous Example -2 : Draw the graph of ` |x|+|y|= 1`

Step 1 : We know the graph for `y =1-x` is shown as

Step 2 : `y = 1 - x -> y = 1 - | x |`

Step 3 : `y = 1-|x| -> 1- |x|`.

Clearly above figure represents a square.

Miscellaneous Example -3 : Draw the graph of `|y|= |e^(-|x|)-1/2|`

Step: 1 we know the graph for `y =e^(-x)`.

Step 2 : `y=e^(-x) -> y=e^(-x) -1/2`

Step 3 : `y=e^(-x) -1/2 -> y= e^(-|x|) -1/2`

Step 4 : `y= e^(-|x|) -1/2 -> y=|e^(-|x|) -1/2|`

Step 5 : `y= |e^(-|x|) -1/2 | -> |y| =|e^(-|x|)-1/2|`

Miscellaneous Example : 4 Draw the graph of `|y| = | 1 - | x-1||`

Step 1 : As we know the graph for y = x - 1

`y = (x- 1) -> y =| x- 1|`

`y = |x - 1 | -> y = - | x - 1 |`

`y = - |x-1| -> y = 1 - |x - 1|`

`y = 1 - |x - 1| -> y = |1-|x-1||`

`y = |1 - |x - 1|| -> |y| = |1-|x-1||`

Miscellaneous Example -5 : Draw the graph of ` |x|-|y|= 1`

Step 1 : As the graph for `y = x - 1` is known;

Step 2 : `y=x-1-> y=|x| -1 |`

Step 3 : `y= |x|-1 -> |y|= |x| -1`

Miscellaneous Example -6 : Draw the graph of `|y|= |e^(x)-1|`

Step 1 : As we know the curve for `y = e^x` is shown as;

Step 2 : `y=e^x -> y=e^x-1`

Step 3 : `y=e^x-1 -> y= |e^x-1 | `

Step 4 : `y= | e^x -1 | -> |y|= | e^x-1 |`

Miscellaneous Example - 7 : Draw the graph of `|y| = | log | x| |`

Step 1 : Here `y= log x` is plotted as;

Step 2 : `y= log x -> y = log |x| `

Step 3 : `y = log |x| -> y = | log | x | |`

Step 4 : `y = | log |x | | -> |y |= | log | x | |`

Miscellaneous Example - 8 : Draw the graph of `| y | = | sin x +1/2 |`

Step 1 : Here ; We know that the graph for `y = sin x` is shown as

Step 2 : `y = sin x -> y = sin x + 1/2 `

Step 3 : `y= sin x +1 /2 -> y= | sin x + 1/2 |`

Step 4 : ` y= | sin x + 1/2 | -> |y| =|sin x +1/2 |`

Miscellaneous Example - 9 : Draw the graph of ` | x | + | y | le 2`

Step 1 : Here; `y= 2- x` is plotted as

`y = 2- x ->y= 2- |x|`

`y= 2- |x| -> |y|= 2- |x|`

`|y| + |x| =2 -> |y| + |x| le 2`

Miscellaneous Example - 10 : Draw the graph of `| y | le | 1 +e^(|x|) - e^(-x) |`

Step 1 : Here `y= e^(|x|)-e^(-x)`

`=> y= { tt (( e^x -e^(-x); , x ge 0), (e^(-x)-e^(-x); , x < 0))`

`y= { tt (( e^x-e^(-x); , x ge 0),(0; , x < 0))`

To discuss;

(i) when `x=0=> y=0` (it passes through origin)

(ii) when `y=0 => e^(2x) -1 =0 => x=0`

(iii) `f(-x) =-f(x)`

as `y=f(x)=e^x-e^(-x)`

`=> f(-x) =e^(-x) -e^x=-f(x);`

it shows `y = f(x) = e^x -e^(-x)` is odd function, i.e., symmetric about origin.

(iv) `y=e^x-e^(-x)`

`=> (dy)/(dx) =e^x+e^(-x) = (e^(2x)+1)/(e^x) > 0` forall `x in R`

`:. y` is increasing for all `x`.

(v) `(d^2y)/(dx^2) =e^x-e^(-x)=(e^(2x)-1)/(e^x)`

`=> (d^2 y)/(dx^2) > 0` when `x >0` concave up and increasing.

also, `(d^2y)/(dx^2) < 0` when `x < 0` concave down and increasing from above discussion `y=e^x-e^(-x)` is plotted as in fig. 1

Now, `y= e^(|x|) -e^(-x) = { tt (( e^x-e^(-x); , x ge 0),(0; , x < 0))`

Thus from fig. 2

Plotting of `y= | 1 + e^(|x|) - e^(-x) |`

`y= e^(|x|)-e^(-x) ->y = 1 + e^(|x|)-e^(-x)`

`=> y= { tt ((1 + e^x -e^(-x); , x ge 0), (1 ; , x < 0))`

Thus, the graph for `y= | 1 + e^(|x|) - e^(-x) |`

It is same as `y= 1 + e^(|x|) - e^(-x) ` {as y `ge` 1 for all `x in R`}

Plotting of ` |y| = | 1 + e^(|x|) - e^(-x) |`

Plotting of `y le | 1 + e^(|x|) - e^(-x) |`

From above figure check any point say `(0,0)`

`0 le 1` (True, thus to shade area towards (0, 0).

From given figure shade part represents the area bounded between two curves.