What is the equation of the line tangent to # f(x)=2x^2 + cos(x)# at # x=pi/3#?

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Answer 1 · 2015-12-06T04:33:18

#y-(4pi^2+9)/18=(8pi-3sqrt3)/6(x-pi/3)#

Find #f'(pi/3)#, the slope of the tangent line at the point on #f(x)# where #x=pi/3#.

The point is at #(x,f(pi/3))#.

#f(pi/3)=2(pi/3)^2+cos(pi/3)=(2pi^2)/9+1/2=(4pi^2+9)/18#

#(pi/3,(4pi^2+9)/18)#

Find the derivative.

#f'(x)=4x-sin(x)#

#f'(pi/3)=4(pi/3)-sin(pi/3)=(4pi)/3-sqrt3/2=(8pi-3sqrt3)/6#

Write in point-slope form:

#y-(4pi^2+9)/18=(8pi-3sqrt3)/6(x-pi/3)#