contestada

consider the functions f(x)=3x^2 - 5 and g(x)= √x -5 +2
how can i find
f(5)
g(5)
f(4)
g(4)
describe the domain of f(x)
describe the domain of g(x)
why is the domain of one of these functions more restructive than the other?

Respuesta :

BiologiaMagister

F(5):

[tex]f(5) = 3(5)^{2} - 5[/tex]

[tex]f(5) = 15^{2} - 5[/tex]

[tex]f(5) = 225 - 5[/tex]

[tex]f(5) = 220[/tex]


G(5):

[tex]g(5) = \sqrt{5} - 5 + 2[/tex]

[tex]g(5) = 2,24 - 5 + 2[/tex]

[tex]g(5) = -2,76 + 2[/tex]

[tex]g(5) = -0,76[/tex]


F(4):

[tex]f(4) = 3(4)^{2} - 5[/tex]

[tex]f(4) = 12^{2} - 5[/tex]

[tex]f(4) = 144 - 5[/tex]

[tex]f(4) = 139[/tex]


G(4):

[tex]g(4) = \sqrt{4} - 5 + 2[/tex]

[tex]g(4) = 2 - 5 + 2[/tex]

[tex]g(4) = -3 + 2[/tex]

[tex]g(4) = -1[/tex]



Hope it helped,


BioTeacher101

tramserran

f(x) = 3x² - 5      

f(5) = 3(5)² - 5   →   f(5) = 3(25) - 5   →   f(5) = 75 - 5   →   f(5) = 70

f(4) = 3(4)² - 5   →   f(4) = 3(16) - 5   →   f(4) = 48 - 5   →   f(5) = 43

DOMAIN of f(x): All real numbers (because there are no restrictions on "x")

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g(x) = [tex]\sqrt{x - 5}[/tex] + 2

g(5) = [tex]\sqrt{5 - 5}[/tex] + 2  →  g(5) =  [tex]\sqrt{0}[/tex] + 2   →   g(5) = 2

g(4) = [tex]\sqrt{4 - 5}[/tex] + 2   →  g(4) =  [tex]\sqrt{-1}[/tex] + 2   →   g(4) = no real solutions

DOMAIN of g(x): x ≥ 5    (because you cannot take the square root of a negative number)


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