What is the equation of the normal line of #f(x)= 2(x-2)^4+(x-1)^3+(x-3)^2# at #x=4#?

Answer 1

#78y=-x+4680#

First find the slope of the tangent at the given point by differentiating and then substituting in #x=4#
#f'(x)=8(x-2)^3 +3(x-1)^2 +2(x-3)#
At #x=4# this is #m =8*2^3+3*2^2+2*1 = 64+12+2 = 78#
The slope of the normal line is #-1/m = -1/78#
The equation of the normal is then #y = -1/78x+c#
To calculate #c#, find the value of the original function at the given point and then substitute into the equation of the normal.
#f(4) = 2*2^4+3^3 +1^2 = 32+27+1 = 60#
#:.60=-1/78*4+c#
#c=60+4/78 = 4680/78#
#y=-1/78x+4680/78#
#:.78y=-x+4680#
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Answer 2

To find the equation of the normal line at a given point on a curve, we need to determine the slope of the tangent line at that point and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line at x=4, we first need to find the derivative of the function f(x). Taking the derivative of f(x) = 2(x-2)^4 + (x-1)^3 + (x-3)^2, we get f'(x) = 8(x-2)^3 + 3(x-1)^2 + 2(x-3).

Evaluating f'(x) at x=4, we have f'(4) = 8(4-2)^3 + 3(4-1)^2 + 2(4-3) = 8(2)^3 + 3(3)^2 + 2(1) = 64 + 27 + 2 = 93.

The slope of the tangent line at x=4 is therefore 93.

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line. Thus, the slope of the normal line is -1/93.

Now that we have the slope of the normal line, we can use the point-slope form of a line to find the equation of the normal line. Since the normal line passes through the point (4, f(4)), we substitute x=4 and f(4) into the point-slope form.

Using the given function f(x), we find f(4) = 2(4-2)^4 + (4-1)^3 + (4-3)^2 = 2(2)^4 + (3)^3 + (1)^2 = 32 + 27 + 1 = 60.

Therefore, the equation of the normal line of f(x) = 2(x-2)^4 + (x-1)^3 + (x-3)^2 at x=4 is y - 60 = (-1/93)(x - 4).

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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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