What is the equation of the normal line of #f(x)=5x^4+3x^3+x^2-2x-4# at #x=1#?

Answer 1

#29y+x-88=0#

Differentiate the function to find its gradient. #f'(x)=20x^3+9x^2+2x-2# Use the point #x=1# to find the gradient of the curve at that point. #f'(1)=20+9+2-2=29# The gradient of the normal is the negative reciprocal of the gradient of the tangent at a point. So the gradient of the normal at #x=1# is #m=-1/29# To find the coordinates of the point the normal passes through, substitute #x=1# into #f(x)# #f(1)=5+3+1-2-4=3# So coordinates are #(x_1, y_1)=>(1, 3)# Equation of a line is: #y-y_1=m(x-x_1)# #y-3=-1/29(x-1)#
#29y-87=-(x-1)#
#29y+x-88=0#
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Answer 2

The equation of the normal line of f(x)=5x^4+3x^3+x^2-2x-4 at x=1 is y = -4x + 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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