# Given y=x^3+x , write an equation for the tangents at the points where the slope is 4, what is the smallest slope of the curve, and at what value of x does the curve has the smallest slope.

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The function y = x^3 + x.

The equation for the tangents at the point where the slope is 4 is required.

The slope of the tangent at any point is equal to the value of the derivative y'.

y' = 3x^2+ 1

3x^2 + 1 = 4

=> 3x^2 = 3

=> x^2 = 1

=> x = 1 and x = -1

At x = 1, y = 2 and at x = -1, y = -2

The equation of the tangents are:

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

=> y - 2 = 4x - 4

=> 4x - y - 2 = 0

And `(y + 2)/(x + 1) = 4`

=> y + 2 = 4x + 4

=> 4x - y + 2 = 0

The slope at any point is equal to 3x^2 + 1

This is least when 6x = 0

=> x = 0

**The value of x when the slope is the least is x = 0 and the least value of the slope is 1.**

The expression for the slope of the tangent of a function at any point is given by the first derivative (dy/dx) or y'.

For the given function y = x^3 + x, the expression for the slope is as under:

slope y' = dy/dx = d(x^3+x)/dx

i.e. **y' = 3 * x^2 + 1** .............. **(1)**

If (x1,y1) are the co-ordinates of the point on the given function y=x^3+x, then

the **equation of the tangent **is can be worked out from the expression:

(y-y1) / (x-x1) = y'

or y-y1 = y' * (x-x1)

or ** y = y' * (x-x1) + y1 ** ............... **(2)**

which is the equation of the tangent at any point of the function at point (x1,y1) of the function

Now to find the equation of the tangent where slope (y') of the tangent is equal to 4, we detirmine the co-ordinates of the point(s) (x1,y1) by substituting the value of slope in equation (1) and we get

4 = 3 * x^2 + 1

or 3 * x^2 = 4 - 1 = 3

or x^2 = 3 / 3 = 1

or x = -1,1 which are the values of x1

Now substituting these values of x1 in the given function

y = x^3 + x,

we get the corresponding values of y1 us under:

for x1 = -1, y1 = (-1)^3 + (-1) = -2 and

for x1 = 1, y1 = 1^3 + 1 = 2

Thus the points **(x1,y1)** on the curve where slope of the tangent is 4 are:

**(-1,-2) **and** (1,2)**

By substituting the calculated values of y', x1 & y1 in equation (2), we get the **equations of tangents **where **slope y' = 4** as under.

For point (-1,-2) on the curve, the equation of tangent is:

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

or y = 4 * x + 4 - 2

or** y = 4 * x +2** ...................... **(3)**

and for point (1,2) on the curve, the equation of tangent is:

y = 4 * (x - 1) + 2

or **y = 4 * x - 2** ...................... **(4)**

Hence **the equation of tangents where slope is equal to 4** are given at **(3)** and **(4)** above.

The **smallest slope of the curve is at the point where y' = 0 and is zero.**

The value of x where y' = 0 can be determined by substituting the value of y' = 0 in equation (1) i.e. y' = 3 * x^2 + 1

thus 0 = 3 * x^2 +1

or 3 * x^2 = -1 / 3

or x = (-1 / 3)^0.5 and x = -1 * (-1 / 3)^0.5

Both these values are imaginary numbers

hence **the curve does not have any point where the slope = 0**