Question

Which one of the following options is true for the function \(f:\mathbb{R}\to\mathbb{R}\) defined by \(f(x) = \min\{|x|,\, x^2-1\}\), \(x\in\mathbb{R}\)?

• A. f is a discontinuous function in R.
• B. f is continuous and differentiable everywhere in R.
• C. f is continuous and differentiable everywhere except at one point of R.
• D. f is continuous and differentiable everywhere except at two points of R.

Hint:

Find where |x| = x^2 - 1 to split f into two smooth pieces, then compare one-sided derivatives at the crossover points.

Answer 1

The Correct Option is D

Step 1: Find where the two pieces cross. Set \(|x| = x^2 - 1\). Writing \(t = |x| \ge 0\), this becomes \(t^2 - t - 1 = 0\), so \(t = \dfrac{1 \pm \sqrt5}{2}\). Only the positive root is admissible since \(t \ge 0\), giving \(t = \varphi = \dfrac{1+\sqrt5}{2} \approx 1.618\). So the two pieces meet exactly at \(x = \varphi\) and \(x = -\varphi\).

Step 2: Determine which piece is smaller on each region. The sign of \(x^2 - 1 - |x| = t^2 - t - 1\) (with \(t=|x|\)) is negative for \(0 \le t < \varphi\) and positive for \(t > \varphi\), since \(t^2-t-1\) is an upward parabola in \(t\) with its only nonnegative root at \(t=\varphi\). So: \[f(x) = \begin{cases} x^2 - 1, & |x| \le \varphi \\ |x|, & |x| \ge \varphi \end{cases}\]

Step 3: Check continuity. On \(|x| < \varphi\), \(f(x)=x^2-1\) is a polynomial, continuous. On \(|x| > \varphi\), \(f(x) = |x|\) is continuous. At \(x = \pm\varphi\), both formulas agree since \(\varphi^2 - 1 = \varphi\) (because \(\varphi\) satisfies \(\varphi^2 = \varphi + 1\)). So \(f\) is continuous on all of \(\mathbb{R}\); option (A) is ruled out.

Step 4: Check differentiability at x = 0. Near \(x=0\), \(|x| < \varphi\), so \(f(x) = x^2-1\) identically in a neighborhood of \(0\). This is a smooth polynomial, so \(f\) is differentiable at \(x=0\) with \(f'(0)=0\). There is no issue at \(x=0\).

Step 5: Check differentiability at x = phi. For \(x\) slightly less than \(\varphi\), \(f(x) = x^2-1\), so the left derivative at \(\varphi\) is \(2\varphi\). For \(x\) slightly greater than \(\varphi\), \(f(x) = x\), so the right derivative is \(1\). Since \(2\varphi = 1+\sqrt5 \approx 3.236 \ne 1\), the two one-sided derivatives disagree, so \(f\) is not differentiable at \(x=\varphi\).

Step 6: Check differentiability at x = -phi. By an identical argument, the left derivative at \(-\varphi\) (from \(f(x)=-x\) for \(x<-\varphi\)) is \(-1\), and the right derivative (from \(f(x)=x^2-1\) for \(x>-\varphi\)) is \(2(-\varphi) = -2\varphi\). These also disagree, so \(f\) is not differentiable at \(x=-\varphi\).

Step 7: Conclude. f is continuous everywhere but fails to be differentiable at exactly the two points \(x = \varphi\) and \(x=-\varphi\). \[\boxed{\text{f is continuous and differentiable everywhere except at two points of } \mathbb{R}}\]