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If g(x) = 3x + 5, what is g^–1(x)?
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if u mean \[ \large g^{-1}(x) \] then just solve for \(x\) the equation \[ \large y=g(x)=3x+5 \]
YES
solve the equation and tell what u got
X=-1 2/3
no
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u have \[ \large y=3x+5 \] first substract 5 from both sides
x= -5+y/3
u r missing some parenthesis!!!!!!
where
g-1= ( g(x) - 5 )/3
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do u see the difference between \[ \large x=-5+\frac{y}{3} \] (this is what u wrote) and \[ \large x=\frac{-5+y}{3}=(-5+y)/3 \]
which one is correct??
the second one
yes!! do u see how important parentheses are??? they can change completely an expression!!!!!
so \[ \large g^{-1}(y)=\frac{y-5}{3} \]
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