Determine the point at which the graph of the function has a horizontal tangent line. y = 16x^-1 - x²

Solution:

Given, f(x) = 16x^-1 - x²

We have to find the points at which the graph of the function has a horizontal tangent line.

A point on a function will have a horizontal tangent line where the first derivative is zero.

Thus, f’(x) = -16x^-2 - 2x

Now, f’(x) = 0

-16x^-2 - 2x = 0

2x = -16x^-2

Dividing by 2x on both sides,

1 = -8x^-3

1 = -8/x³

x³ = -8

Taking cube root,

x = -2

Put x = -2 in f(x)

f(-2) = 16/-2 - (-2)²

= -8 - 4

= -12

Therefore, the point is (-2, -12).

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Determine the point at which the graph of the function has a horizontal tangent line. y = 16x^-1 - x²

Summary:

The points at which the graph of the function has a horizontal tangent line. f(x) = 16x^-1 - x² is (-2, -12).