What is the equation of the tangent line of #f(x)=sqrt(x)# at #x=4#?

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Answer 1 · 2018-07-27T18:16:09

#x-4y+4=0#

Setting #x=4# in the given function, we get y-coordinate of point as follows

#y=f(4)#

#=\sqrt4#

#=2#

hence the point has coordinates #(4, 2)#

Now, the slope of tangent #dy/dx# at any point to the function #f(x)=\sqrtx# is given as

#dy/dx=f'(x)#

#=d/dx(\sqrtx)#

#=1/{2\sqrtx}#

Now,the slope #m# of tangent at the given point #(4, 2)# is given as

#m=f'(4)#

#=1/{2\sqrt4}#

#=1/4#

hence the equation of tangent with slope #m=1/4# & at the point #(x_1, y_1)\equiv(4, 2)# is given by following formula

#y-y_1=m(x-x_1)#

#y-2=1/4(x-4)#

#4y-8=x-4#

#x-4y+4=0#