What is the equation of the line tangent to #f(x)=sin x + cos^2 x # at #x=0#?

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Answer 1 · 2016-12-12T11:34:33

The equation of the tangent line is:

#y(x) = 1+x#

The equation of the line tangent to the curve #y=f(x)# for #x=barx # is:

#y(x) = f(bar x) + f'(bar x)(x- barx)#

Here:

#f(x) =sin x +cos^2x#
#f'(x) =cosx -2sinxcosx = cosx -sin(2x)#

and for #bar x =0#

#f(barx) = 1#
#f'(barx) = 1#

so the equation of the tangent line is:

#y(x) = 1+x#

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Answer 2 · 2016-12-12T11:35:00

#y=x+1#

At #x=0#, #y = f(0) = 1#. (The y-intercept is #1#.)

#f'(x) = cosx-2sinxcosx#, so at #x=0#, the slope of the tangent line is

#m = f'(0) = 1#.

The line with slope #1# and y-intercept #1# has equation

#y=x+1#.