Evaluate the function at the specified value of the independent variable and simplify. $f ( y ) = \left\{ \begin{array} { l l }
y , & y \leq - 1 \
y ^ { 2 } - 2 y , & - 1 \leq y \leq 1 \
y ^ { 3 } - 2 y ^ { 2 } , & y 1
\end{array} \right.$
$f \left( \frac { 4 } { 5 } \right)$
A) $\frac { 4 } { 15 }$
B) $\frac { 4 } { 5 }$
C) $- \frac { 96 } { 125 }$
D) $- \frac { 24 } { 25 }$
E) $- \frac { 64 } { 125 }$

Question

Evaluate the function at the specified value of the independent variable and simplify. $f ( y ) = \left\{ \begin{array} { l l } 
y , & y \leq - 1 \
y ^ { 2 } - 2 y , & - 1 \leq y \leq 1 \
y ^ { 3 } - 2 y ^ { 2 } , & y > 1
\end{array} \right.$
$f \left( \frac { 4 } { 5 } \right)$
A) $\frac { 4 } { 15 }$
B) $\frac { 4 } { 5 }$
C) $- \frac { 96 } { 125 }$
D) $- \frac { 24 } { 25 }$
E) $- \frac { 64 } { 125 }$

Quizplus · Accepted Answer

Since $\frac{4}{5}$ falls within the range $-1 \leq y \leq 1$, we use the second piece of the function, $y^2 - 2y$. Substituting $\frac{4}{5}$ into this equation gives $\left(\frac{4}{5}\right)^2 - 2\left(\frac{4}{5}\right) = \frac{16}{25} - \frac{8}{5} = \frac{16}{25} - \frac{40}{25} = -\frac{24}{25}$.

Evaluate the Function at the Specified Value of the Independent