Answer
$${\text{Relative maximum at }}\left( {2,2.21429} \right)$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \arctan x - \arctan \left( {x - 4} \right) \cr
& {\text{*Calculate the first derivative}} \cr
& f'\left( x \right) = \frac{1}{{1 + {x^2}}} - \frac{1}{{1 + {{\left( {x - 4} \right)}^2}}} \cr
& {\text{Set }}f'\left( x \right) = 0 \cr
& \frac{1}{{1 + {x^2}}} - \frac{1}{{1 + {{\left( {x - 4} \right)}^2}}} = 0 \cr
& \frac{1}{{1 + {x^2}}} = \frac{1}{{1 + {{\left( {x - 4} \right)}^2}}} \cr
& 1 + {x^2} = 1 + {\left( {x - 4} \right)^2} \cr
& 1 + {x^2} = 1 + {x^2} - 8x + 16 \cr
& 1 + {x^2} - 1 - {x^2} = - 8x + 16 \cr
& 0 = - 8x + 16 \cr
& x = 2 \cr
& \cr
& {\text{Evaluate }}f'\left( 1 \right){\text{ and }}f'\left( 3 \right) \cr
& f'\left( 1 \right) = \frac{1}{{1 + {{\left( 1 \right)}^2}}} - \frac{1}{{1 + {{\left( {1 - 4} \right)}^2}}} = \frac{2}{5} > 0 \cr
& f'\left( 3 \right) = \frac{1}{{1 + {{\left( 3 \right)}^2}}} - \frac{1}{{1 + {{\left( {1 - 3} \right)}^2}}} = - \frac{2}{5} > 0 \cr
& {\text{The derivative changes from positive to negative at }}x = 2 \cr
& {\text{By the first derivative test, there is a relative maximum at }} \cr
& \left( {2,f\left( 2 \right)} \right) \cr
& f\left( 2 \right) = \arctan \left( 2 \right) - \arctan \left( {2 - 4} \right) \cr
& f\left( 2 \right) \approx 2.21429 \cr
& {\text{Relative maximum at }}\left( {2,2.21429} \right) \cr} $$
