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Sagot :
we can check each option using the limit number f(x)=3 and a number out of the range(4)
must have a value for the limit number but a mistake when we use 4
A.
f(x)=3
[tex]\begin{gathered} 3=2^x+3 \\ 3-3=2^x \\ 0=2^x \end{gathered}[/tex]any value of x will result in 0
then A is wront because dont have a solution for the limit number
B.
f(x)=3
[tex]\begin{gathered} 3=3x+8 \\ 3-8=3x \\ -5=3x \\ x=-\frac{5}{3} \end{gathered}[/tex]has a solution for f(x)=3
f(x)=4
[tex]\begin{gathered} 4=3x+8 \\ 4-8=3x \\ -4=3x \\ x=-\frac{4}{3} \end{gathered}[/tex]has a solution for a value out of the range then option B is wrong
C.
f(x)=3
[tex]\begin{gathered} 3=-x^2+3 \\ 3-3=-x^2 \\ 0=-x^2 \\ -x=\sqrt[]{0} \\ -x=0 \\ x=0 \end{gathered}[/tex]has a solution for f(x)=3
f(x)=4
[tex]\begin{gathered} 4=-x^2+3 \\ 4-3=-x^2 \\ 1=-x^2 \\ x^2=-1 \\ x=\sqrt[]{-1} \end{gathered}[/tex]root of negative number doesnt have real solution , then this fuction dont have solution for a number out of the range
Option C is RIGHT because has solution on the limit number f(x)=3 but not a solution for a number out of range
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