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

Answer 1

Emma Aldridge

#" "y=1/7x-15/7#

The gradient of the line is found by differentiation

Let #" "y=-2x^3-3x^2+5x-2# .......................(1)

Gradient #= (dy)/(dx)= -6x^2-6x+5#

Thus the gradient of #f(x)# at #x=1# is

#(dy)/(dx)=-6(1)^2-6(1)+5 = -7# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The gradient of the line that is normal to #f(x)# is #" "(-1)xx(dx)/(dy)= +1/7#

So we have

#" "y=1/7x+c#......................(2)

We have a known point #P_("(x,y)") ->P_("(1,y)")#

We can determine the value of y by #f(x)->f(1)#

#" "f(1)->y=-2(1)^3-3(1)^2+5(1)-2#

#" "y= -2#

We now have #P_("(x,y)") ->P_("(1,-2)")#

So by substitution in equation (2) #" "-2=1/7(1)+c#......................(2)

#" "c= -15/7#

Giving

#" "y=1/7x-15/7......................(2_a)#

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Answer 2

William Abdalla

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