For $(r \circ q)(x)$, we have:
$(r \circ q)(x) = r(q(x)) = r(2x + 1)$
Now, we substitute the expression for $r(x)$:
$(r \circ q)(x) = -2(2x + 1)^2 - 2$
For $(q \circ r)(x)$, we have:
$(q \circ r)(x) = q(r(x)) = q(-2x^2 - 2)$
Now, we substitute the expression
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