For positive real numbers \(S\) and \(K\), the function \(H_K(S)\) is defined as: \[ H_K(S) = \max(S-K, 0) \] The max function is defined as: \[ \max(a,b) = \begin{cases} a, & \text{when } a > b \\ b, & \text{when } a \leq b \end{cases} \] The graph below shows the plot of a function \(N(S)\) versus \(S\). \(N(S)\) can be expressed as ______.
Show Hint
A single \(\max(S-K,0)\) hinge produces one kink; two kinks in the graph mean you need two hinge terms whose slopes cancel appropriately after \(S=20\).
Step 1: Instead of proving the identity region by region, test each candidate formula at two convenient sample points read directly from the graph and eliminate any formula that fails a numeric check.
Step 2: Pick sample points from the graph. From the plot, \(N(15)=5\) (halfway up the ramp between \(S=10\) and \(S=20\)) and \(N(25)=10\) (on the flat plateau after \(S=20\)).
Step 3: Evaluate option (A) at these points. \(H_{10}(15)-H_{20}(15) = (15-10) - 0 = 5\), matching \(N(15)=5\). \(H_{10}(25)-H_{20}(25) = (25-10)-(25-20) = 15-5=10\), matching \(N(25)=10\). Option (A) passes both checks.
Step 4: Evaluate option (B) at the same points. \(H_{10}(15)-2H_{20}(15) = 5 - 0 = 5\), which happens to pass the first check, but \(H_{10}(25)-2H_{20}(25) = 15-2(5)=5\), which does not match \(N(25)=10\). Option (B) fails.
Step 5: Evaluate option (C) at \(S=15\). \(-H_{10}(15)+H_{20}(15) = -5+0=-5\), which does not match \(N(15)=5\) at all (wrong sign). Option (C) fails immediately.
Step 6: Evaluate option (D) at \(S=15\). \(H_{15}(15)-H_{20}(15) = 0-0=0\), which does not match \(N(15)=5\). Option (D) fails.
Step 7: Conclude. Only option (A) survives both numeric checks, confirming the same answer found by the region by region derivation. \[ \boxed{N(S) = H_{10}(S) - H_{20}(S)} \]
