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𝑓(𝑥) = 𝑥 − 5, 𝑔(𝑥) = 𝑥 2 − 𝑥 − 6, ℎ(𝑥) = √4𝑥 + 3, 𝑝(𝑥) = 9𝑥 2 − 25.
1. x-intercepts are points on graph where y=0.
a) 𝑓(𝑥) = 0, 𝑥 − 5 = 0, 𝑥 = 5. The only x-intercept is (𝟓, 𝟎).
b) 𝑔(𝑥) = 𝑥 2 − 𝑥 − 6 = 0, guess 𝑥1 = 3 and by Viet’s theorem 𝑥2 = −2.
The intercepts are (−𝟐, 𝟎), (𝟑, 𝟎).
2. y-intercepts are points on graph where x=0.
a) 𝑓(0) = −5, the y-intercept is (𝟎, −𝟓).
b) 𝑔(0) = −6, the y-intercept is (𝟎, −𝟔).
c) ℎ(0) = √3, the y-intercept is (𝟎, √𝟑).
3.
a) 𝑓(2) = 2 − 5 = −𝟑,
b) 𝑔(−3) = (−3)2 − (−3) − 6 = 𝟔,
c) ℎ(3) = √12 + 3 = √𝟏𝟓,
d) 𝑔(𝑥 − 2) = (𝑥 − 2)2 − (𝑥 − 2) − 6 = 𝑥 2 − 4𝑥 + 4 − 𝑥 + 2 − 6 = 𝒙𝟐 − 𝟓𝒙,
e) 𝑓(𝑔(2)) = 𝑓(22 − 2 − 6) = 𝑓(−4) = −4 − 5 = −𝟗,
f) 𝑝(ℎ(−2)) does not exist because ℎ is not defined at 𝑥 = −2 (4 ∙ (−2) + 3 = −5 < 0),
g) (𝑔 + 𝑝)(𝑥) = 𝑔(𝑥) + 𝑝(𝑥) = 𝑥 2 − 𝑥 − 6 + 9𝑥 2 − 25 = 𝟏𝟎𝒙𝟐 − 𝒙 − 𝟑𝟏,
h) (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥) = 𝑥 − 5 − 𝑥 2 + 𝑥 + 6 = −𝒙𝟐 + 𝟐𝒙 + 𝟏,
i) (𝑓 ∙ 𝑝)(𝑥) = 𝑓(𝑥)𝑝(𝑥) = (𝑥 − 5)(9𝑥 2 − 25) = 𝟗𝒙𝟑 − 𝟒𝟓𝒙𝟐 − 𝟐𝟓𝒙 + 𝟏𝟐𝟓.
4.
a) (−∞, ∞),
3
3
b) 4𝑥 + 3 ≥ 0, 𝑥 ≥ − 4 , [− 4 , ∞),
𝑝
𝑝(𝑥)
c) 𝑓 (𝑥) = 𝑓(𝑥), the only restriction is 𝑓(𝑥) ≠ 0, 𝑥 ≠ 5, 𝑥 ∈ (−∞, 𝟓) ∪ (𝟓, ∞).
3
𝟑
d) 𝑝 works with any 𝑥 so the only restriction is 4𝑥 + 3 ≥ 0, 𝑥 ≥ − 4 , [− 𝟒 , ∞).
5.
1
a) 𝑓(𝑥) = (2𝑥 − 4)−3 = (2𝑥−4)3 , the only restriction is 2𝑥 − 4 ≠ 0, 𝑥 ≠ 2, 𝑥 ∈ (−∞, 𝟐) ∪ (𝟐, ∞),
b) 𝑓(𝑥) =
√𝑥−3
.
𝑥−8
It must be 𝑥 − 3 ≥ 0, 𝑥 − 8 ≠ 0, i.e. 𝑥 ≥ 3, 𝑥 ≠ 8, 𝑥 ∈ [𝟑, 𝟖) ∪ (𝟖, ∞),
6.
a) |6𝑥 − 5| ≤ 4.
5
1
For 6𝑥 − 5 ≤ 0, 𝑥 ≤ 6 we have −(6𝑥 − 5) ≤ 4, −6𝑥 ≤ −1, 𝑥 �...