How do you find the points where the tangent line is horizontal given #y=16x^-1-x^2#?

Question

Isabella Ackerman · Accepted Answer

The point at which the tangent line is horizontal is #(-2, -12)#. To find the points at which the tangent line is horizontal, we have to find where the slope of the function is 0 because a horizontal line's slope is 0. #d/dxy = d/dx(16x^-1 - x^2)#
#d/dxy = -16x^-2 - 2x# That's your derivative. Now set it equal to 0 and solve for x to find the x values at which the tangent line is horizontal to given function. #0 = -16x^-2 - 2x#
#2x = -16/x^2#
#2x^3 = -16#
#x^3 = -8#
#x = -2# We now know that the tangent line is horizontal when #x = -2# Now plug in #-2# for x in the original function to find the y value of the point we're looking for. #y = 16(-2)^-1 - (-2)^2 = -8 - 4 = -12#  The point at which the tangent line is horizontal is #(-2, -12)#. You can confirm this by graphing the function and checking if the tangent line at the point would be horizontal: graph{(16x^(-1)) - (x^2) [-32.13, 23, -21.36, 6.24]}

Axel Alvarez · Answer

undefined To find the points where the tangent line is horizontal for the equation y = 16x^(-1) - x^2, we need to find the values of x where the derivative of y with respect to x equals zero.

First, we find the derivative of y with respect to x:
dy/dx = -16x^(-2) - 2x

Next, we set the derivative equal to zero and solve for x:
-16x^(-2) - 2x = 0

To simplify the equation, we multiply through by -x^2:
16 + 2x^3 = 0

Rearranging the equation, we have:
2x^3 = -16

Dividing both sides by 2, we get:
x^3 = -8

Taking the cube root of both sides, we find:
x = -2

Therefore, the point where the tangent line is horizontal is when x = -2.