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?

Relax

Respuesta :

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

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)