What is the slope of the line normal to the tangent line of #f(x) = x^3-sqrtx # at # x= 3 #?

Answer 1

The slope of tangent can be found by taking the first derivative. The slope of the normal is the negative reciprocal of the slope of the tangent.

#f(x) = x^3-sqrt(x)# Slope of tangent # = f'(x) = 3x^2 - 1/(2sqrt(x))# Slope of tangent at #x=3# is #f'(3) =3(3)^2-1/(2sqrt(3))#
Slope of tangent #= 27 - 1/(2sqrt(3)#
Simplifying we get #(54sqrt(3) - 1)/(2sqrt(3))#

The slope of normal would be the negative reciprocal of this value.

Slope of normal #= (-2sqrt(3))/(54sqrt(3)-1)#
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Answer 2

The slope of the line normal to the tangent line of f(x) = x^3 - sqrt(x) at x = 3 is -1/18.

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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.

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.

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.

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.

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