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If f(x) = (1/3)x + 9, which statement is always true?

(1) f(x) < 0 (3) If x < 0, then f(x) < 0.
(2) f(x) > 0 (4) If x > 0, then f(x) > 0.

Respuesta :

naǫ
[tex]f(x)=\frac{1}{3}x+9 \\ \\ f(x)\ \textless \ 0 \\ \frac{1}{3}x+9\ \textless \ 0 \\ \frac{1}{3}x\ \textless \ -9 \\ x\ \textless \ -27 \\ \\ f(x)\ \textgreater \ 0 \\ \frac{1}{3}x+9\ \textgreater \ 0 \\ \frac{1}{3}x\ \textgreater \ -9 \\ x\ \textgreater \ -27[/tex]

f(x)<0 for x<-27, f(x)>0 for x>-27.
Therefore, if x<0 then f(x) can be either less or greater than 0. If x>0 then f(x) is greater than 0.

Statement (4) is always true.
[tex]1) f(x) \ \textless \ 0[/tex]

=> We do not know the value of x. Therefore, we cannot be sure. Not sure.

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[tex]2) f(x) \ \textgreater \ 0[/tex]

=> The same way. X is unknown, it is not always true.

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3)  If x < 0, then f(x) < 0

=> It doesn't provide in each case.
 
Let's take two examples:

◘ Suppose that; x = -30

[tex] \frac{1}{3} x + 9 = \frac{1}{3} (-30) + 9 [/tex]

[tex]= (-10) + 9 = -1[/tex]

Okay, it provided.

◘ But, x= -3

[tex] \frac{1}{3} x + 9 = \frac{1}{3} (-3) + 9 = 8[/tex]

x <0  but  f(x) > 0

False.

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4) x > 0, then f(x) > 0

That's true !

Examples:

x= 6

[tex] \frac{1}{3} x + 9 = \frac{1}{3}. 6 + 9 = 2 + 9 = 11[/tex]

f(6) = 11

Namely, 

if   x>0  ,    f(x) >0 

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Answer= 4

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