How do you find the points where the tangent line is horizontal given y= 16x^-1 - x²

Solution:

Given, y= 16x^-1 - x²

To find the points at which the tangent line is horizontal, we have to find where the slope of the function is zero because the slope of a horizontal line is zero.

dy/dx = d(16x^-1 - x²)

dy/dx = -16x^-2 - 2x

To find the slope of the horizontal line, set dy/dx to zero.

-16x^-2 - 2x = 0

It can be written as

2x = -16/x²

By cross multiplication

2x³ = -16

Dividing both sides by 2

x³ = -8

Taking cube root,

x = -2

This implies that the tangent line is horizontal at x = -2

Put the value of x in the given function,

y =16(-2)^-1 - (-2)²

y = -8 - 4

y = -12

Therefore, the point at which the tangent line is horizontal is (-2, -12).