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What is the equation of the tangent line of #f(x)=sqrt(x^2-x+7# at #x=3#?

Answer 1

#2sqrt(13)y = 5x +11#

The slope of of the tangent line is found by differentiating the function. The tangent line is given by #y = mx +c# where #m# is the slope, and the constant can be found from the coordinates of the point on the original curve.

#f'(x) = 1/2(x^2-x+7)^(-1/2)*(2x - 1)# #f'(x)=(2x-1)/(2sqrt(x^2-x+7))# At #x=3# #f'(x) = (2*3-1)/(2sqrt(9-3+7)# #f'(x) = 5/(2sqrt(13)# At the same point #f(x) = sqrt(9-3+7) = sqrt(13)#

Therefore for the tangent line, #sqrt(13) = (5/(2sqrt(13)))*3+c# #c = sqrt(13) - 15/(2*sqrt(13)) = (26-15)/(2*sqrt(13)) =11/(2sqrt(13))# The tangent line is therefore #y = 5/(2sqrt(13))x +11/(2sqrt(13)#

#2sqrt(13)y = 5x +11#

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Answer 2

Charles Arocha

The equation of the tangent line of f(x) = sqrt(x^2 - x + 7) at x = 3 is y = 2x - 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.