What is the equation of the line normal to #f(x)=x ^3-3x^2 # at #x=4#?

Answer 1

#y = frac(1)(24) x - frac(23)(6)#

We have: #f(x) = x^(3) - 3 x^(2)#
First, let's find the #y#-coordinate corresponding to the given value of #x#:
#Rightarrow f(4) = (4)^(3) - 3 cdot (4)^(2)#
#Rightarrow f(4) = 64 - 48#
#therefore f(4) = 16#
So we must find the equation of the normal line at the point #(4, 16)#.
Then, let's differentiate #f(x)#:
#Rightarrow f'(x) = 3 x^(2) - 6 x#
For #x = 4#, the slope of the line is:
#Rightarrow f'(4) = 3 cdot (4)^(2) - 6 cdot (4)#
#Rightarrow f'(4) = 48 - 24#
#therefore f'(4) = 24#

The gradient of the normal line is the reciprocal of this value:

#Rightarrow "Gradient of normal line" = (f'(4))^(- 1)#
#therefore "Gradient of normal line" = frac(1)(24)#

Now, let's express the equation of the normal line in point-slope form:

#Rightarrow y - y_(1) = m (x - x_(1))#
#Rightarrow y - 16 = frac(1)(24) (x - 4)#
#Rightarrow y - 4 = frac(1)(24) x - frac(1)(6)#
#therefore y = frac(1)(24) x - frac(23)(6)#
Therefore, the equation of the normal line is #y = frac(1)(24) x - frac(23)(6)#.
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Answer 2

To find the equation of the line normal to the function f(x) = x^3 - 3x^2 at x = 4, we need to determine the slope of the tangent line at that point and then find the negative reciprocal to obtain the slope of the normal line.

To find the slope of the tangent line, we take the derivative of the function f(x). The derivative of f(x) = x^3 - 3x^2 is f'(x) = 3x^2 - 6x.

Next, we substitute x = 4 into the derivative to find the slope of the tangent line at x = 4. f'(4) = 3(4)^2 - 6(4) = 48 - 24 = 24.

Since the slope of the normal line is the negative reciprocal of the slope of the tangent line, we have: Slope of normal line = -1/24.

Now, we have the slope of the normal line and the point (4, f(4)) = (4, 4^3 - 3(4)^2) = (4, 16 - 48) = (4, -32).

Using the point-slope form of a line, we can write the equation of the line normal to f(x) at x = 4: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point. Therefore, the equation of the line normal to f(x) = x^3 - 3x^2 at x = 4 is: y - (-32) = (-1/24)(x - 4).

Simplifying the equation, we get: y + 32 = (-1/24)x + 1/6.

Rearranging the equation, we have: y = (-1/24)x - 31/6.

Thus, the equation of the line normal to f(x) = x^3 - 3x^2 at x = 4 is y = (-1/24)x - 31/6.

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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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