Question

suppose that the functions q and r are defined as follows.
q(x) = 2x + 2
r(x) = x² - 1
find the following.
(q ∘ r)(-1) =
(r ∘ q)(-1) =

Explanation:

Step1: Find \((q \circ r)(-1)\)

First, recall that \((q \circ r)(x)=q(r(x))\). So we need to find \(r(-1)\) first, then plug that into \(q(x)\).

Find \(r(-1)\): \(r(x)=x^2 - 1\), so \(r(-1)=(-1)^2 - 1 = 1 - 1 = 0\).

Now plug \(r(-1)=0\) into \(q(x)\): \(q(x)=2x + 2\), so \(q(0)=2(0)+2 = 0 + 2 = 2\). Thus, \((q \circ r)(-1)=2\).

Step2: Find \((r \circ q)(-1)\)

Recall that \((r \circ q)(x)=r(q(x))\). So we need to find \(q(-1)\) first, then plug that into \(r(x)\).

Find \(q(-1)\): \(q(x)=2x + 2\), so \(q(-1)=2(-1)+2 = -2 + 2 = 0\).

Now plug \(q(-1)=0\) into \(r(x)\): \(r(x)=x^2 - 1\), so \(r(0)=0^2 - 1 = 0 - 1 = -1\). Thus, \((r \circ q)(-1)=-1\).

Answer:

\((q \circ r)(-1)=\boxed{2}\)
\((r \circ q)(-1)=\boxed{-1}\)