Question

the function f(x) is invertible. find $f^{-1}(3)$.
$f^{-1}(3)=\square$

Explanation:

Step1: Recall inverse function definition

The inverse function \( f^{-1}(y) \) gives the \( x \) such that \( f(x) = y \). So, to find \( f^{-1}(3) \), we need to find \( x \) where \( f(x) = 3 \).

Step2: Analyze the graph of \( f(x) \)

Look at the graph of \( f(x) \). We need to find the \( x \)-value corresponding to \( y = 3 \). By examining the grid, when \( y = 3 \), we check the \( x \)-coordinate on the curve of \( f(x) \). From the graph, when \( f(x) = 3 \), the \( x \)-value is \( 1 \) (since the curve passes through a point where \( y = 3 \) and \( x = 1 \) approximately? Wait, no, wait. Wait, let's re-examine. Wait, the graph is a curve starting from near \( y \)-axis, decreasing. Wait, when \( x = 0 \), \( f(0) = 3 \)? Wait no, at \( x = 0 \), the \( y \)-value is 3? Wait, no, looking at the graph: the \( y \)-axis at \( x = 0 \), the curve is at \( y = 3 \)? Wait, no, the grid: each square is 1 unit. So when \( y = 3 \), what's \( x \)? Wait, maybe I made a mistake. Wait, the key is: \( f^{-1}(3) \) is the \( x \) such that \( f(x) = 3 \). So we need to find \( x \) where \( f(x) = 3 \). Looking at the graph, when \( y = 3 \), the \( x \)-coordinate is \( 1 \)? Wait, no, let's check again. Wait, the graph: at \( x = 1 \), what's \( y \)? Wait, maybe the graph is a function like \( f(x) = 3 \cdot (1/2)^x \) or something, but from the grid, when \( y = 3 \), the \( x \) is \( 1 \)? Wait, no, maybe I messed up. Wait, actually, the correct approach: the inverse function swaps \( x \) and \( y \). So the point \( (a, b) \) on \( f(x) \) corresponds to \( (b, a) \) on \( f^{-1}(x) \). So to find \( f^{-1}(3) \), we find the \( x \) where \( f(x) = 3 \), i.e., find the \( x \) such that \( (x, 3) \) is on \( f(x) \). Looking at the graph, when \( y = 3 \), the \( x \)-value is \( 1 \)? Wait, no, let's look at the graph again. The curve is in the first quadrant, decreasing. At \( x = 0 \), \( y = 3 \)? Wait, at \( x = 0 \), the \( y \)-coordinate is 3? Then \( f(0) = 3 \), so \( f^{-1}(3) = 0 \)? Wait, that makes sense. Wait, maybe I misread the graph. Let's check: the \( y \)-axis (x=0) intersects the curve at y=3? So \( f(0) = 3 \), so \( f^{-1}(3) = 0 \)? Wait, no, wait, the graph: at x=0, the y-value is 3? Let's see the grid: the vertical axis is y, horizontal is x. Each square is 1 unit. So at x=0, the curve is at y=3. So \( f(0) = 3 \), so \( f^{-1}(3) = 0 \)? Wait, that seems correct. Because if \( f(0) = 3 \), then \( f^{-1}(3) = 0 \).

Wait, maybe I made a mistake earlier. Let's confirm: the definition of inverse function is \( f(f^{-1}(y)) = y \) and \( f^{-1}(f(x)) = x \). So if \( f(0) = 3 \), then \( f^{-1}(3) = 0 \). So looking at the graph, at x=0, y=3. So that's the point (0, 3) on f(x). Therefore, \( f(0) = 3 \), so \( f^{-1}(3) = 0 \).

Step3: Confirm

So since \( f(0) = 3 \), by the definition of inverse function, \( f^{-1}(3) = 0 \).

Answer:

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