What is the equation of the line tangent to # f(x)=sinpi/x # at # x=1 #?

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Answer 1 · 2015-11-25T13:51:24

#y=0#

#sinpi=0#, so, for all #x != 0#, we have #f(x)=0#.

Therefore, #f'(1)=0#. (By definition or by properties/rules for differentiation.)

At the point where #x=1#, we get #y=f(1)=0#,

and the equation of the line with slope #m=0# through #(1,0)# is #y=0#

Answer 2 · 2015-11-25T16:05:16

If the question should have been for #f(x)=sin(pi/x)#, then the equation of the tangent line is #y=pix-pi#

For #f(x)=sin(pi/x)#, we get #f(1)=sinpi=0#

and
#f'(x) = cos(pi/x) [d/dx(pi/x)] = cos(pi/x) [d/dx(pi x^-1)]#

# = cos(pi/x) [-pi x^-2] = -pi/x^2 cos (pi/x)#

So, #f'(1) = -picospi = pi#

The line through #(1,0)# with slope #m=pi# is

#y=pix - pi#