given the following piecewise function
F(x)={x+11 for -9¡Üx<-3
2 for -3¡Üx¡Ü2
2x-6 for 2<x¡Ü5
{--this symbol is to be around all three of the lines.
Find:
a. the domain
b. the range
c. the intercepts
d. if f continious on its domain. If
not, state where f is discontiniuos
e. graph the function
Please show work. Thanks!!
domain is all x where f is defined: [-9,5]
range: (-2,8]
intercepts: (3,0)
discontinuous at x=-3,2
To find the domain of the given piecewise function, we need to consider the x-values for which the function is defined.
a. Domain: In the given piecewise function, we have specified the ranges of x-values for which the function is defined. So, we can say that the domain is:
Domain: [-9, -3) ∪ [-3, 2] ∪ (2, 5]
b. To find the range of the function, we need to determine the set of all possible y-values that the function can take on.
Range: Looking at each piece of the function:
1. For the first piece, which is F(x) = x + 11, we see that as x goes from -9 to -3 (inclusive), the function values increase from x + 11 = -9 + 11 = 2 to x + 11 = -3 + 11 = 8. So, the range for this piece is [2, 8].
2. For the second piece, which is F(x) = 2, the function is constant. So, the range for this piece is {2}.
3. For the third piece, which is F(x) = 2x - 6, as x goes from 2 to 5 (exclusive), the function values increase from 2(2) - 6 = -2 to 2(5) - 6 = 4. So, the range for this piece is [-2, 4].
Combining all the ranges, we get:
Range: [-2, 8] ∪ {2}
c. To find the intercepts of the function, we need to find the x-intercepts and y-intercepts.
x-intercepts: These are the points where the function crosses or touches the x-axis, which means the y-values are zero.
To find the x-intercepts, set the y-value to zero for each piece of the function and solve for x.
1. For the first piece, x + 11 = 0, x = -11.
2. For the second piece, 2 = 0 (which has no solution).
3. For the third piece, 2x - 6 = 0, x = 3.
So, the x-intercepts are x = -11 and x = 3.
y-intercept: This is the point where the function crosses or touches the y-axis, which means the x-value is zero.
To find the y-intercept, substitute x = 0 for each piece of the function and evaluate the y-value.
1. For the first piece, F(0) = 0 + 11 = 11.
2. For the second piece, F(0) = 2.
3. For the third piece, F(0) = 2(0) - 6 = -6.
So, the y-intercept is y = 11, and the function also passes through the point (0, 2) and (0, -6) as per the respective pieces.
d. To determine if the function is continuous on its domain, we need to check if the function values approach the same value as x approaches the endpoints of each interval and if there are no jumps or gaps in the graph.
1. At x = -3, the function has a jump from 2 to (-3 + 11) = 8.
2. At x = 2, the function has a jump from 2 to (2)(2) - 6 = -2.
Thus, the function is discontinuous at x = -3 and x = 2.
e. To graph the function, plot the different sections of the function on the x-y plane based on the domain and range, and ensure that the jumps and gaps are indicated.