1. **State the problem:** We have two functions $f$ and $g$ given by their graphs. We need to find the domain and range of $f$, evaluate expressions involving $f$ and $g$, find inverse values, and determine if $g$ is invertible. 2. **Domain and range of $f$:** - The domain is the set of all $x$ values for which $f(x)$ is defined. - The range is the set of all $y$ values that $f(x)$ takes. 3. **Evaluate $(3f - 5g)(4)$:** - Use the formula $(3f - 5g)(4) = 3f(4) - 5g(4)$. - Find $f(4)$ and $g(4)$ from the graphs. 4. **Evaluate $(f \, o \, g)(-1)$:** - This means $f(g(-1))$. - Find $g(-1)$, then find $f$ at that value. 5. **Evaluate $(g \, o \, f)(-4)$:** - This means $g(f(-4))$. - Find $f(-4)$, then find $g$ at that value. 6. **Find $f^{-1}(2)$:** - Find the $x$ such that $f(x) = 2$. 7. **Is $g$ invertible?** - A function is invertible if it is one-to-one (passes the horizontal line test). --- **Step-by-step solution:** 1. **Domain of $f$:** From the graph, $f$ is defined from $x = -3$ (filled point) to $x = 5$ (filled point), so domain is $[-3,5]$. 2. **Range of $f$:** The lowest $y$ value is $-2$ (at $x=5$), highest is $3$ (at $x=0$), so range is $[-2,3]$. 3. **Evaluate $(3f - 5g)(4)$:** - From graph, $f(4) = -1$ (filled point near $x=4$). - From graph, $g(4) = -2$ (filled point at $x=4$). - Calculate: $$ (3f - 5g)(4) = 3 \times (-1) - 5 \times (-2) = -3 + 10 = 7 $$ 4. **Evaluate $(f \circ g)(-1)$:** - From graph, $g(-1) = 1$ (on left side of $g$'s graph). - Then $f(1) = 2$ (from $f$ graph). - So: $$ (f \circ g)(-1) = f(g(-1)) = f(1) = 2 $$ 5. **Evaluate $(g \circ f)(-4)$:** - From graph, $f(-4) = 1$. - Then $g(1) = 1$. - So: $$ (g \circ f)(-4) = g(f(-4)) = g(1) = 1 $$ 6. **Find $f^{-1}(2)$:** - Find $x$ such that $f(x) = 2$. - From graph, $f(1) = 2$, so: $$ f^{-1}(2) = 1 $$ 7. **Is $g$ invertible?** - $g$ is an inverted V shape with a peak at $x=0$. - It fails the horizontal line test because some $y$ values correspond to two $x$ values. - Therefore, $g$ is not invertible. **Final answers:** - Domain of $f$: $[-3,5]$ - Range of $f$: $[-2,3]$ - $(3f - 5g)(4) = 7$ - $(f \circ g)(-1) = 2$ - $(g \circ f)(-4) = 1$ - $f^{-1}(2) = 1$ - $g$ is not invertible because it is not one-to-one.