Determine g(-1) for g(t)=(t^2-1)(t+4)
a)-2
b)-6
c)3
d)0

It is C because Since {{g(x)}={x^3+2}}g(x)=x3+2g, left parenthesis, x, right parenthesis, equals, x, start superscript, 3, end superscript, plus, 2, we can substitute {x^3+2}x3+2x, start superscript, 3, end superscript, plus, 2 in for {g(x)}g(x)g, left parenthesis, x, right parenthesis.\begin{aligned}{f(g(x))}&=3(g(x))-1 &=3({x^3+2})-1 &=3x^3+6-1 &=3x^3+5 \end{aligned}f(g(x))=3(g(x))−1=3(x3+2)−1=3x3+6−1=3x3+5

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