What is the equation of the line tangent to # f(x)=x^2-sqrt(e^x-3x) # at # x=0#?

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Answer 1 · 2017-02-10T12:59:24

#x+y+1=0#
See the tangent-inclusive Socratic dented-for-discontinuity graph.

graph{(x^2-sqrt(e^x-3x)-y)(x-y-1)(x^2+(y+1)^2-.01)=0 [-10, 10, -5, 5]}

To make f real, #e^s>=3x#.

Nearly, for #x in (0.5, 1.5), f is not real. See the graph below, wherein.

The graph for #e^x# is below the graph for #3x#

graph{(e^x-y)(y-3x)=00 [0.6 1.5, -5, 5]}

f = -1, at x =1.,

So, the point of contact of the tangent is #P( 0, -1 )#.

#f' = 2x-(e^x-3)/(2sqrt(e^x-3x))=1#, at P.

So, the equation to the tangent at #P( 0, -1 )# is

#y-(-1)=(1)(x-0)#, giving

#x-y-1=0#