What is the equation of the line tangent to # f(x)=sqrt(e^x-x) # at # x=1#?

Question

1640 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-30T05:48:46

2y=#sqrt(e-1)x +sqrt(e-1)#
OR y=#sqrt(e-1)x/2+sqrt(e-1)/2#

f(x)= #sqrt(e^x-x)

First derivative at x=1 is the slope of the tangent to f(x) at x=1

f'(x)= #(e^x-1)/(2sqrt(e^x-x))(using [chain rule](http://socratic.org/calculus/basic-differentiation-rules/chain-rule) ) #
f'(1)=#(e-1)/(2sqrt(e-1))#

f'(1) =#(e-1)/2#

slope =#(e-1)/2# and points (1,f(1) ) ; f(1)=#sqrt(e-1)#

using equation of line in point slope form

y-y1=m(x-x1) ; m=slope=#(e-1)/2# ;x1=1 and y1=#sqrt(e-1)#

y-#sqrt(e-1)#=#(e-1)/2#(x-1)

2y-2#sqrt(e-1)#=#sqrt(e-1)#x-#sqrt(e-1)#

2y=#sqrt(e-1)#x+#sqrt(e-1)#

y=#sqrt(e-1)x/2+sqrt(e-1)/2#