Question

The figure below shows the graph of a function \(f(x)\). How many solutions does the equation \(f(f(x))=15\) have?

 

• A. 5
• B. 6
• C. 7
• D. 8
• E. cannot be determined from the given graph

Hint:

To solve \(f(f(x))=k\) from a graph: first find all \(t\) with \(f(t)=k\), then for each such \(t\) count how many \(x\) satisfy \(f(x)=t\); add the counts.

Answer 1

The Correct Option is C

Step 1: Find all \(t\) such that \(f(t)=15\).
From the graph, the horizontal line \(y=15\) meets the curve at exactly two \(x\)-values: one at the top of the “peak’’ near \(x=4\), and another on the far right rising branch near \(x\approx 11\).
Hence the set \(S=\{t:\ f(t)=15\}=\{4,\;11\}\).
Step 2: Count the solutions of \(f(x)=t\) for each \(t\in S\).
\(\bullet\) For \(t=4\): the line \(y=4\) intersects the graph 4 times — once on the left branch (for negative \(x\)) and once on each of the three right-side branches (the increasing piece \(0\!\to\!4\), the decreasing piece \(4\!\to\!9\), and the final rising piece \(9\!\to\!12\)). Hence \(\#\{x:\ f(x)=4\}=4\).
\(\bullet\) For \(t=11\): the line \(y=11\) intersects the graph 3 times — once on the increasing piece \(0\!\to\!4\), once on the decreasing piece \(4\!\to\!9\), and once on the rightmost rising piece; the left branch never reaches \(y=11\). Thus \(\#\{x:\ f(x)=11\}=3\).
Step 3: Each solution of \(f(x)=4\) or \(f(x)=11\) is a solution of \(f(f(x))=15\). Therefore the total number of solutions is \[ 4+3=7. \] \(\boxed{7}\).