find equations of both lines that are tangent to the curve y=1 + x^3 and are parallel to the line 12x - y = 1?

Answer 1

The line
y = 12 x – 1
has a slope 12.

Tangent lines have equations
y = 12 x + c₁
and
y = 12 x + c₂
where c₁ and c₂ are constants to be found.

Now let us find the points, where our cubic parabola
y = x³ + 1
has a derivative equal to 12.

The derivative function is
y‘ = 2 x².

Let us solve
12 = 2 x².

Solutions are
x₁ = √6
and
x₂ = -√6.

Corresponding values of our cubic parabola are
y₁ = 6√6 + 1
y₂ = -6√6 + 1.

Now let us solve equations
6√6 + 1 = 12√6 + c₁
and
-6√6 + 1 = -12√6 + c₂.

Solutions are
c₁ = 1 - 6√6
and
c₂ = 1 + 6√6.

Answer:

Equations of tangent lines are
y = 12 x + 1 - 6√6
and
y = 12 x + 1 + 6√6.

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