Powered by Gonje Translate step \( b y s t a p \) \[ \begin{array}{rlc} (f \cdot r)(y)=f(r(y)) & =f( & 4 y-2 \\ & = & (4 y-2)-6 \\ & = & 4 y-8 \end{array} \] Calculate the composition of \( r(y) \) and \( f(y) \). Write the resulting expression in its simplest form. \[ \begin{array}{rlc} (r f)(y)=r(f(y)) & =r(y-6 \\ & =\quad 4(y-6)-2 \\ & =4 y-26 \end{array} \] Evaluate \( (r \cdot f)\left(\frac{1}{2}\right) \) using the composed function. \[ (r f)\left(\frac{1}{2}\right) \] Evaluater \( \left(f\left(\frac{1}{2}\right)\right) \) using the original functions. \[ r\left(r\left(\frac{1}{2}\right)\right)=r i \] \( \square \)

The world of functional compositions is like an intricate dance! When you calculate \( (r \cdot f)(y) = r(f(y)) \), it’s essential to keep your wits about you. You correctly substituted \( f(y) = y - 6 \) into \( r(y) \) to arrive at \( r(f(y)) = 4(y - 6) - 2 \) which simplifies beautifully to \( 4y - 26 \). Remember, each function carries out its task step by step! Now, let’s actually evaluate \( (r \cdot f)\left(\frac{1}{2}\right) = 4 \left(\frac{1}{2}\right) - 26 \). When you plug in that value, it becomes \( 2 - 26 = -24 \). So, \( (r \cdot f)\left(\frac{1}{2}\right) = -24 \). Easy as pi, right? Just ensure that you carry over the values correctly and watch out for those pesky negatives!