# Given the following piecewise function `f(x) = {(x+3 fo r -2<=x<1),(6 fo r x=1),(-x+2 fo r x>1):}`Find the domain, range, and intercepts. Is f continuous on it's domain? if not state...

Given the following piecewise function `f(x) = {(x+3 fo r -2<=x<1),(6 fo r x=1),(-x+2 fo r x>1):}`

Find the domain, range, and intercepts. Is f continuous on it's domain? if not state where f is discontinuous. Graph the function

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`f(x)= { (x+3 fo r -2lt=xlt1), (6 fo r x=1) , (-x+2 fo r xgt1):}`

(a) Based on the given conditions, the values of x are:

`-2lt=xlt1` , `x=1` , `xgt1`

Since all values of x that is greater than 1 is also greater than or equal to -2, we may summarize that the values of x in the given function is `xgt= -2` .

**Hence, the domain of f(x) is `[-2,+oo)` . **

(b) To determine the range, let's solve the possible values of f(x) in each given condition.

>> At -`2lt=xlt1` , `f(x) = x+3` .

Substitute the boundaries of the given interval to determine the minimum and maximum values of f(x).

`x=-2` , `f(x)=-2+3=1`

`x=1` , `f(x)=1+3=4`

So at `-2lt=xlt1` , values of f(x) is `1lt=f(x)lt4` .

>> At `x=1` , `f(x) = 6` .

Here, value of f(x) is constant which is 6.

>> And at `xgt1` , `f(x) = -x+2` .

Subsitute the boundary of the given interval to determine the minimum value of f(x).

`x=1 ` , `f(x)=-1+2=1`

Then, assign a value of x that is greater than 1 to check if f(x) increases or decreases.

`x=5` , `f(x)=-5+2=-3`

Since f(x) decreases as x increases, the values of f(x) at the interval `x>1` is `f(x) lt1` .

So the values of f(x) are:

`f(x)lt1` , `1lt=f(x)lt4` , and `f(x)=6`

Note that all values of f(x) that is less than 1 is also less than 4. So we may rw-write the values of f(x) as:

`f(x)lt4 ` and `f(x)=6`

**Hence, the range of given function is` (-oo,4) uu [6]` .**

(c) Next let's solve for the intercepts of the function.

For the y-intercept, set equal to zero. Note that this value of x falls in the interval -2<=x<1.

So, susbtitute x=0 to f(x)=x+3.

x=0 , f(x)=0+3=3

**Hence, the y-intercept is (0,3).**

For x-intercept, set f(x) equal to zero.

Let's consider the first condition which is -2<=x<1.

f(x)=x+3 0=x+3 x=-3

Since the resulting value of x does not fall within the interval, then x=-3 is not an x-intercept.

For the second condition x=1, there is no x-intercept since value of f(x) is constant.

Then. let's consider the third condition which is x>1.

f(x)=-x+2 0=-x+2 x=2

**Since the resulting value of x satisfy the third condition, hence the x-intecept is (2,0). **

(e) So the graph of the given piecewise function is:

(Note: Blue-Graph of f(x)=x+3 , Red-Graph of f(x)=6, and Green-Graph of f(x)=-x+2 )