Find the tangents to the curve y= x^3 + x at the points where the slope is 4.?

Question

What is the smallest slope of the curve. At what value of x does the curve have this slope?

I want a detailed Xplanation so that i can understand it, please.

So if you're just going to giving me the answer, thanx, but no thanx.

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Answer 1

y = x^3 + x ...(1)

dy / dx = 3x^2 + 1

As the slope of the required tangents is 4, this is the value of dy / dx at the points of contact.

3x^2 + 1 = 4

3x^2 = 3

x^2 = 1

x = +/- 1.

Substituting these values of x in (1) gives the corresponding y co-ordinates:

When x = 1, y = 1^3 + 1 = 2.

When x = - 1, y = (- 1)^3 + (- 1) = - 2.

The tangents are at (1, 2) and (- 1, - 2).

You now have the gradient (4) and the point of contact for each tangent.

The equation of a line passing through (1, 2) with gradient 4 is:

y - 2 = 4(x - 1)

y = 4x - 2.

The equation of a line passing through (- 1, - 2) with gradient 4 is:

y - (- 2) = 4(x - (- 1))

y + 2 = 4x + 1

y = 4x - 1.

The smallest slope of the curve is the minimum value of dy / dx.

That is the minimum value of 3x^2 + 1.

As the minimum value of x^2 is 0 (a square is never negative), the minimum value of 3x^2 + 1 is:

3 * 0 + 1 = 1

when x = 0.