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Question
find g(a - 1) when g(x) = 3x - 3.
a. 3a - 3
b. \frac{1}{3}a - 3
c. 3a - 6
d. 3a + 1
Step1: Substitute \( x = a - 1 \) into \( g(x) \)
We know that \( g(x)=3x - 3 \), so to find \( g(a - 1) \), we replace \( x \) with \( a - 1 \) in the function. So we get \( g(a - 1)=3(a - 1)-3 \).
Step2: Expand and simplify the expression
First, expand \( 3(a - 1) \) using the distributive property \( 3\times a-3\times1 = 3a-3 \). Then we have \( g(a - 1)=3a-3 - 3 \). Combining like terms, \( -3-3=-6 \), so \( g(a - 1)=3a - 6 \).
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C. \( 3a - 6 \)