3.3 Practice - Piecewise Functions Find the value of the given function at the indicated domain value.
-X2 - 5x + 2, x < 1
2x2 - 2x + 1,
xs-6
g(x) = x3 - 5x, 1 < x < 11
h) =
3x – x3 -3 3x - 16,
X > 11
(2x - (x - 100
x > 1
1. g(1) =
2. g(11) =
3. h(5) =
4 h(-10) =
5. g(-1) =
6. h(0) =
7. g(20) =
8. h(-4)=

Step-by-step explanation:

1. g(1) = (1)³ - 5(1) = - 4

2. At x = 11 function g(x) is not defined,

3. h(5) = 2(5) - | 5 - 10 | = 5

4 h( - 10) = 2( - 10)² - 2(- 10) + 1 = 221

5. g(- 1) = - (- 1)² - 5(- 1) + 2 = - 1 + 5 + 2 = 6

6. h(0) = 3(0) - (0)³ = 0

7. g(20) = - √[3(20) - 16] = - 2√11

8. h(- 4) = not defined

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)