Question

The local minimum value of the function \[ f(x)=2x-3\tan^{-1}x \] is

• A. \(3\tan^{-1}\!\left(\dfrac1{\sqrt2}\right)-\sqrt2\)
• B. \(\sqrt2-3\tan^{-1}\!\left(\dfrac1{\sqrt2}\right)\)
• C. \(3\tan^{-1}(\sqrt2)-\dfrac1{\sqrt2}\)
• D. \(\dfrac1{\sqrt2}-3\tan^{-1}(\sqrt2)\)

Hint:

For local extrema, \[ \boxed{f'(x)=0} \] gives the critical points, while \[ \boxed{ f''(x)>0 } \] indicates a local minimum and \[ \boxed{ f''(x)<0 } \] indicates a local maximum.

Answer 1

The Correct Option is B

Step 1: Find the critical points. Given, \[ f(x)=2x-3\tan^{-1}x. \] Differentiate: \[ f'(x)=2-\frac{3}{1+x^2}. \] Setting \[ f'(x)=0, \] we get \[ 2=\frac{3}{1+x^2}. \] Hence, \[ 1+x^2=\frac32, \] \[ x^2=\frac12, \] \[ x=\pm\frac1{\sqrt2}. \]

Step 2: Determine the nature of the critical points. Differentiate once more: \[ f''(x)=\frac{6x}{(1+x^2)^2}. \] At \[ x=\frac1{\sqrt2}, \] \[ f''\!\left(\frac1{\sqrt2}\right)>0, \] so the function has a local minimum.

Step 3: Find the minimum value. Substituting \[ x=\frac1{\sqrt2}, \] \[ f\!\left(\frac1{\sqrt2}\right) = \frac2{\sqrt2} - 3\tan^{-1}\!\left(\frac1{\sqrt2}\right). \] Since \[ \frac2{\sqrt2}=\sqrt2, \] we obtain \[ \boxed{ f_{\min} = \sqrt2 - 3\tan^{-1}\!\left(\frac1{\sqrt2}\right). } \] Hence, the correct option is \(\boxed{(B)}\).