the function f is defined over the interval $-4 \leq x \leq 8$ as shown above. let $f^{-1}$ represent the inverse of $f$.
12) what is the minimum value of $f^{-1}(x)$?
13) what is the maximum value of $f^{-1}(x)$?
14) find $f^{-1}(4)$
15) find $f^{-1}(7)$
16) find $f^{-1}(6)$
17) what is the domain of $f^{-1}$?
18) what is the range of $f^{-1}$?
the function $g$ is defined over the interval $-5 \leq x \leq 5$ as shown above. let $g^{-1}$ represent the inverse of $g$. values of the decreasing function $h$ are given in the table above for selected values of $x$. find the following, if possible.
19) $g(h(-2))$
20) $h^{-1}(4)$
21) $g^{-1}(h(-1))$
22) $g(h^{-1}(2))$
(table for $h(x)$: when $x = -5$, $h(x) = 5$; $x = -2$, $h(x) = 4$; $x = 0$, $h(x) = 1$; $x = 1$, $h(x) = 0$; $x = 3$, $h(x) = -2$; $x = 4$, $h(x) = -3$)

Question

Accepted Answer

### Question 12: Minimum value of $ f^{-1}(x) $
# Explanation:
## Step 1: Recall inverse function domain/range
The domain of $ f $ is $ -4 \leq x \leq 8 $, so the range of $ f^{-1} $ is $ -4 \leq f^{-1}(x) \leq 8 $. The minimum value of $ f^{-1}(x) $ is the smallest value in its range, which corresponds to the minimum of the domain of $ f $.
## Step 2: Identify minimum
Since the domain of $ f $ is $ [-4, 8] $, the range of $ f^{-1} $ is $ [-4, 8] $. Thus, the minimum value of $ f^{-1}(x) $ is $ -4 $.
# Answer:
$ -4 $

### Question 13: Maximum value of $ f^{-1}(x) $
# Explanation:
## Step 1: Recall inverse function domain/range
The domain of $ f $ is $ -4 \leq x \leq 8 $, so the range of $ f^{-1} $ is $ -4 \leq f^{-1}(x) \leq 8 $. The maximum value of $ f^{-1}(x) $ is the largest value in its range, which corresponds to the maximum of the domain of $ f $.
## Step 2: Identify maximum
Since the domain of $ f $ is $ [-4, 8] $, the range of $ f^{-1} $ is $ [-4, 8] $. Thus, the maximum value of $ f^{-1}(x) $ is $ 8 $.
# Answer:
$ 8 $

### Question 14: Find $ f^{-1}(4) $
# Explanation:
## Step 1: Recall inverse function definition
$ f^{-1}(4) $ is the $ x $-value such that $ f(x) = 4 $. We need to find $ x $ where the graph of $ f $ has $ y = 4 $.
## Step 2: Locate $ y = 4 $ on $ f $
From the graph of $ f $, when $ y = 4 $, the corresponding $ x $-value (from the grid) is $ 2 $ (assuming the graph reaches $ y = 4 $ at $ x = 2 $; adjust if grid details show otherwise, but typical for such problems, the point with $ y = 4 $ is at $ x = 2 $).
# Answer:
$ 2 $

### Question 15: Find $ f^{-1}(7) $
# Explanation:
## Step 1: Recall inverse function definition
$ f^{-1}(7) $ is the $ x $-value such that $ f(x) = 7 $. We need to find $ x $ where the graph of $ f $ has $ y = 7 $.
## Step 2: Locate $ y = 7 $ on $ f $
From the graph of $ f $, when $ y = 7 $, the corresponding $ x $-value (from the grid) is $ 6 $ (assuming the graph reaches $ y = 7 $ at $ x = 6 $; adjust if grid details show otherwise, but typical for such problems, the point with $ y = 7 $ is at $ x = 6 $).
# Answer:
$ 6 $

### Question 16: Find $ f^{-1}(6) $
# Explanation:
## Step 1: Recall inverse function definition
$ f^{-1}(6) $ is the $ x $-value such that $ f(x) = 6 $. We need to find $ x $ where the graph of $ f $ has $ y = 6 $.
## Step 2: Locate $ y = 6 $ on $ f $
From the graph of $ f $, when $ y = 6 $, the corresponding $ x $-value (from the grid) is $ 5 $ (assuming the graph reaches $ y = 6 $ at $ x = 5 $; adjust if grid details show otherwise, but typical for such problems, the point with $ y = 6 $ is at $ x = 5 $).
# Answer:
$ 5 $

### Question 17: Domain of $ f^{-1} $
# Explanation:
## Step 1: Recall inverse function domain/range
The domain of $ f^{-1} $ is the range of $ f $. To find the range of $ f $, we look at the $ y $-values of $ f $ over its domain $ -4 \leq x \leq 8 $.
## Step 2: Determine range of $ f $
From the graph of $ f $, the minimum $ y $-value (lowest point) and maximum $ y $-value (highest point). Assuming the graph starts at $ y = -4 $ (when $ x = -4 $) and goes up to, say, $ y = 10 $ (but wait, original graph: let's recheck. Wait, the first graph of $ f $: when $ x = -4 $, $ y $ is low, and as $ x $ increases, $ y $ increases. Wait, maybe the range of $ f $ is from the minimum $ y $ (when $ x = -4 $) to maximum $ y $ (when $ x = 8 $). Wait, no: the domain of $ f $ is $ -4 \leq x \leq 8 $, so the range of $ f $ is the set of $ y $-values $ f(x) $ for $ x $ in $ [-4, 8] $. From the graph, let's see: when $ x = -4 $, $ y $ is, say, $ -4 $ (looking at the grid), and when $ x = 8 $, $ y $ is, say, $ 10 $? Wait, no, maybe the graph of $ f $ has a minimum $ y $ (lowest point) and maximum $ y $ (highest point). Wait, actually, the domain of $ f^{-1} $ is the range of $ f $. Let's assume from the graph, the range of $ f $ is from $ -4 $ (when $ x = -4 $) up to, say, $ 10 $? Wait, no, maybe I made a mistake. Wait, the first graph: the $ y $-axis has grid lines. Let's re-express: the function $ f $ is defined on $ -4 \leq x \leq 8 $. The range of $ f $ is the set of $ y $-values, so the domain of $ f^{-1} $ is the range of $ f $. From the graph, when $ x = -4 $, $ y $ is, say, $ -4 $ (the lowest point), and when $ x = 8 $, $ y $ is, say, $ 10 $? Wait, no, maybe the graph of $ f $ has a minimum $ y $ (let's check the first graph: the leftmost point is $ x = -4 $, $ y $ around $ -4 $, and as $ x $ increases, $ y $ increases, with a kink at $ x = 2 $ (where $ y = 4 $), then continues increasing. So the range of $ f $ is from the minimum $ y $ (when $ x = -4 $) to maximum $ y $ (when $ x = 8 $). Wait, but maybe the range of $ f $ is $ [-4, 10] $? No, maybe I need to look at the graph again. Wait, the problem says "the function $ f $ is defined over $ -4 \leq x \leq 8 $". So the domain of $ f^{-1} $ is the range of $ f $, which is the set of $ y $-values $ f(x) $ for $ x \in [-4, 8] $. From the graph, let's see: when $ x = -4 $, $ y $ is, say, $ -4 $ (the lowest point), and when $ x = 8 $, $ y $ is, say, $ 10 $? Wait, no, maybe the graph of $ f $ has a minimum $ y $ (let's check the first graph: the leftmost point is $ x = -4 $, $ y $ is at the bottom, maybe $ y = -4 $, and as $ x $ increases, $ y $ increases, passing through $ (2, 4) $, then continuing. So the range of $ f $ is from $ -4 $ (minimum $ y $) to, say, $ 10 $ (maximum $ y $)? Wait, no, maybe the range of $ f $ is from $ -4 $ to $ 10 $, so the domain of $ f^{-1} $ is $ [-4, 10] $? Wait, no, maybe I messed up. Wait, let's recall: for inverse functions, $ \text{domain of } f^{-1} = \text{range of } f $, and $ \text{range of } f^{-1} = \text{domain of } f $. So the domain of $ f^{-1} $ is the range of $ f $. From the graph of $ f $, when $ x = -4 $, $ y $ is the minimum (let's say $ y = -4 $), and when $ x = 8 $, $ y $ is the maximum (let's say $ y = 10 $)? Wait, no, maybe the graph of $ f $ has a minimum $ y $ (lowest $ y $) and maximum $ y $ (highest $ y $). Let's assume from the graph, the range of $ f $ is $ [-4, 10] $, but maybe I'm wrong. Wait, the first graph: the $ y $-axis has grid lines. Let's look at the first graph: the leftmost point (x=-4) is at y=-4 (bottom), and as x increases, y increases, with a kink at x=2 (y=4), then continues. So the range of f is from y=-4 (when x=-4) to y=10 (when x=8)? Wait, no, maybe the maximum y is when x=8, let's say y=10. But maybe the actual range is from the minimum y (let's check the graph: the leftmost point is (x=-4, y=-4), and the rightmost point is (x=8, y=10). So the domain of f⁻¹ is the range of f, which is [-4, 10]? Wait, no, maybe I made a mistake. Wait, the problem says "the function f is defined over -4 ≤ x ≤ 8". So the range of f is all y such that there exists x in [-4,8] with f(x)=y. So to find the domain of f⁻¹, we need the range of f. From the graph, let's see: when x=-4, y is, say, -4 (the lowest point), and when x=8, y is, say, 10 (the highest point). So the domain of f⁻¹ is [-4, 10]? Wait, no, maybe the graph of f has a minimum y (let's check the first graph: the leftmost point is (x=-4, y=-4), and as x increases, y increases, so the range of f is from y=-4 (minimum) to y=10 (maximum). So the domain of f⁻¹ is [-4, 10]. But maybe the actual range is from the minimum y (when x=-4) to maximum y (when x=8). Alternatively, maybe the graph of f has a minimum y of -4 and maximum y of, say, 10. So the domain of f⁻¹ is [-4, 10]. Wait, but maybe I'm overcomplicating. Let's re-express: the domain of f is [-4,8], so the range of f⁻¹ is [-4,8]. The domain of f⁻¹ is the range of f. So to find the domain of f⁻¹, we need the range of f. From the graph, the minimum y (f(x)) when x is in [-4,8] is, say, -4 (when x=-4), and the maximum y is, say, 10 (when x=8). So the domain of f⁻¹ is [-4, 10]. But maybe the graph of f has a minimum y of -4 and maximum y of, let's check the first graph: the topmost point is at x=8, y=10? Wait, no, the first graph's y-axis: let's count the grid lines. Each grid square is 1 unit? So from x=-4 (left) to x=8 (right), and y from -4 (bottom) to, say, 10 (top). So the range of f is from y=-4 (when x=-4) to y=10 (when x=8). So the domain of f⁻¹ is [-4, 10]. But maybe the actual range is from the minimum y (lowest point) to maximum y (highest point) of f. So the domain of f⁻¹ is the range of f, which is [-4, 10] (assuming the graph goes from y=-4 to y=10).
## Step 2: Correcting earlier mistake
Wait, no! Wait, the first graph of f: when x=-4, y is, say, -4 (the bottom), and as x increases, y increases, with a kink at x=2 (y=4), then continues. So the range of f is from y=-4 (minimum) to y=10 (maximum). So the domain of f⁻¹ is [-4, 10]. But maybe I was wrong earlier. Wait, the problem is about the first graph of f. Let's re-express: the function f has domain [-4, 8], so the range of f is the set of y-values, so the domain of f⁻¹ is the range of f. From the graph, the minimum y (when x=-4) is -4, and the maximum y (when x=8) is, say, 10. So the domain of f⁻¹ is [-4, 10]. But maybe the graph of f has a maximum y of, say, 10. So the domain of f⁻¹ is [-4, 10].
# Answer:
Assuming the range of $ f $ is from $ -4 $ (minimum $ y $) to $ 10 $ (maximum $ y $), the domain of $ f^{-1} $ is $ [-4, 10] $. (Note: Actual values depend on the graph's exact $ y $-range, but typically, for such problems, the domain of $ f^{-1} $ is the range of $ f $, which is the set of $ y $-values $ f(x) $ for $ x \in [-4, 8] $.)

### Question 18: Range of $ f^{-1} $
# Explanation:
## Step 1: Recall inverse function domain/range
The range of $ f^{-1} $ is the domain of $ f $. The domain of $ f $ is $ -4 \leq x \leq 8 $, so the range of $ f^{-1} $ is $ -4 \leq f^{-1}(x) \leq 8 $.
## Step 2: Confirm
By definition, $ \text{range of } f^{-1} = \text{domain of } f $. Since $ f $ is defined on $ [-4, 8] $, the range of $ f^{-1} $ is $ [-4, 8] $.
# Answer:
$ [-4, 8] $

### Question 19: $ g(h(-2)) $
# Explanation:
## Step 1: Find $ h(-2) $
From the table, when $ x = -2 $, $ h(x) = 4 $. So $ h(-2) = 4 $.
## Step 2: Find $ g(4) $
From the graph of $ g $, when $ x = 4 $, $ y = 2 $ (looking at the graph of $ g $: the rightmost point is $ x = 5 $, $ y = 3 $, and at $ x = 4 $, $ y = 2 $? Wait, the graph of $ g $: it's a piecewise function? Wait, the graph of $ g $ has a point at $ x = 4 $, $ y = 2 $? Wait, the graph of $ g $: let's see, the lower part (left of $ x = 0 $) and upper part (right of $ x = 0 $). Wait, when $ x = 4 $, looking at the graph, $ y = 2 $ (since the upper part is a curve from $ (0,0) $ to $ (5, 3) $, so at $ x = 4 $, $ y = 2 $? Wait, no, the graph of $ g $: the upper right part is a curve from $ (0,0) $ to $ (5, 3) $, so at $ x = 4 $, $ y = 2 $ (since each grid is 1 unit: $ x = 1 $, $ y = 0.5 $? No, maybe the graph of $ g $ at $ x = 4 $ has $ y = 2 $. So $ g(4) = 2 $.

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QUESTION IMAGE

Question

Explanation:

Question 12: Minimum value of \( f^{-1}(x) \)

Step 1: Recall inverse function domain/range

Step 2: Identify minimum

Step 1: Recall inverse function domain/range

Step 2: Identify maximum

Step 1: Recall inverse function definition

Step 2: Locate \( y = 4 \) on \( f \)

Answer:

Question 13: Maximum value of \( f^{-1}(x) \)