Consider the line y=4 x -1 and the point P=(2,0). (a) Write the formula for a function d(x) that describes the distance between the point P and a point (x,y) on the line. You final answer should only involve the variable x. Then d(x) = √(4−x)2(4x−1)2 (b) d'(x)= (c) The critical number is x= . (d) The closest point on the line to P is ( , ).

Answer:a) d(x)=[tex]\sqrt{17x^{2} -12x+5}[/tex]b)d'(x)=[tex]\frac{17x-6}{\sqrt{17x^{2} -12x+5} }[/tex]c)The critical point is x=[tex]\frac{6}{17}[/tex]d)Closest point is ([tex]\frac{6}{17}[/tex],[tex]\frac{7}{17}[/tex]Step-by-step explanation:We are given the line [tex]y=4x-1[/tex]Let a point Q([tex]x,y[/tex]) lie on the line.Point P is given as P(2,0)By distance formula, we have the distance D between any two pointsA([tex]x_{1},y_{1}[/tex]) and B([tex]x_{2},y_{2}[/tex]) asD=[tex]\sqrt{(x_{1}-x_{2})^2 + (y_{1}-y_2)^2}[/tex]Thus,d(x)=[tex]\sqrt{(x-2)^2+(y-0)^2}[/tex]But we have, [tex]y=4x-1[/tex]So,d(x)=[tex]\sqrt{(x-2)^2+(4x-1)^2}[/tex]Expanding,d(x)=[tex]\sqrt{17x^2-12x+5}[/tex]  - - - (a)Now,d'(x)= [tex]\frac{\frac{d}{dx} (17x^2-12x+5)}{2(\sqrt{17x^2-12x+5}) }[/tex]i.e.d'(x)=[tex]\frac{17x-6}{\sqrt{17x^{2} -12x+5} }[/tex] - - - (b)Now, the critical point is where d'(x)=0⇒ [tex]\frac{17x-6}{\sqrt{17x^{2} -12x+5} }[/tex] =0⇒[tex]x=\frac{6}{17}[/tex]    - - - (c) Now, The closest point on the given line to point P is the one for which d(x) is minimum i.e. d'(x)=0⇒[tex]x=\frac{6}{17}[/tex]as [tex]y=4x-1[/tex]⇒y=[tex]\frac{7}{17}[/tex]So, closest point is ([tex]\frac{6}{17},\frac{7}{17}[/tex])   - - -(d)