# It is given that `f(x) = 4x, 0<=x<1` `f(x) = (3-x)^2, 1<=x<=3` and `f(-x) = -f(x)` It is also known that `f(x) = f(x+6)` for all real values of x. (i) Show that `f(4) = -1` (ii)...

It is given that

`f(x) = 4x, 0<=x<1`

`f(x) = (3-x)^2, 1<=x<=3`

and `f(-x) = -f(x)`

It is also known that `f(x) = f(x+6)` for all real values of x.

(i) Show that `f(4) = -1`

(ii) Sketch the graph of `y=f(x)` for `-6<=x<=9`

Thanks!

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### 1 Answer

The function f(x) is defined as

`f(x) = 4x for 0<=x<1`

`f(x) = (3 - x)^2` for `1<= x<=3`

and `f(-x) = -f(x)`

Also, `f(x) = f(x + 6)` for all real values of x.

To determine the value of f(4):

4 = 6 - 2 = -(2 - 6)

`f(4) = f(-(2-6))`

As f(-x) = f(x), we have

-f(2 - 6)

Now f(x) = f(x+6)

= -f(2 - 6 + 6)

= -f(2)

For 1<=x<=3, f(x) = (3 - x)^2, this gives:

-(3 - 2)^2

= -1^2

= -1

To sketch the graph of y = f(x) for -6<=x<=9 first determine the value of f(x) for all the values of x in the given domain.

[-6,9] = [-6,-3]U[-3, -1]U(-1, 0]U[0,1)U[1, 3]U[3,6]U[6,9]

For `x in [-6, -3]`

`y = f(x) = f(x+6) = (3 - x-6)^2 = (-x - 3)^2`

For `x in [-3, -1]`

`f(x) = -f(x) = -(3 - x)^2`

For `x in (-1, 0]`

`y = -f(x) = -4*x`

For `x in [0,1)`

`y = 4*x`

For `x in [1, 3]`

`y = (3 - x)^2`

For `x in [3, 6]`

`y = f(-(-x))` `= -f(-x)` `= -f(-x + 6)` `= -(3 - (-x+6))^2` `= -(3 + x - 6)^2` `= -(x-3)^2`

For `x in [6,7)`

`y = f(-(-x)) = -f(-x+6) = -4*(6-x)`

For `x in [7,9]`

`y = f(-(-x))` ` = -f(-x)` ` = -f(-x+6)` `= -(-f(x - 6))` `= (3 - (x - 6)^2` `= (-x + 3)^2`

The required graph is: