For #f(x)=4/(x-1) # what is the equation of the tangent line at #x=0#?
1 Answer
Jul 1, 2018
Explanation:
#"we require the gradient of the tangent and a point on it"#
#m_("tangent")=f'(x)" at x = 0"#
#f(x)=4/(x-1)=4(x-1)^-1#
#"differentiate using the "color(blue)"chain rule"#
#f'(x)=-4(x-1)^-2=-4/(x-1)^2#
#f'(0)=-4/1=-4#
#"and "f(0)=4/(-1)=-4rArr(0,-4)#
#y+4=-4x#
#y=-4x-4larrcolor(red)"equation of tangent"#
