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Find the equation of the tangent and the normal to the given curve at the indicated point: y = x4 – 6x3 + 13x2 – 10x + 5 at the point where x = 1

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It is given that

y = x4 – 6x3 + 13x2 – 10x + 5

By substituting the value of x = 1

y = (1)4 – 6(1)3 + 13(1)2 – 10(1) + 5

On further calculation

y = 1 – 6 + 13 – 10 + 5 = 3

Point of contact is (1, 3)

Consider y = x4 – 6x3 + 13x2 – 10x + 5 as the equation of curve

By differentiating w.r.t. x

dy/dx = 4x3 – 18x2 + 26x – 10

So we get

(dy/dx)(1, 3) = 4(1)3 – 18(1)2 + 26(1) – 10 = 2

Here the required equation of tangent is

y – y1 = m(x – x1)

By substituting the values

y – 3 = 2x – 2

On further calculation

2x – y – 2 + 3 = 0

So we get

2x – y + 1 = 0

Here the required equation of normal i

y – y1 = -1/m(x – x1)

By substituting the values

y – 3 = -1/2(x – 1)

On further calculation

2y – 6 = – x + 1

So we get

2y – 6 + x – 1 = 0

x + 2y – 7 = 0

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