Question

Let f(x) x3 - 4x2. Find the point(s) on the graph of f where the tangent line is horizontal.

(x,y)=______ (smaller x-value)
(x,y)=_______ (larger x-value)

Answer 1

The points on the graph of f(x) = x^3 - 4x^2 where the tangent line is horizontal are found by setting the derivative f'(x) = 3x^2 - 8x to zero, yielding x-values of 0 and 8/3. The corresponding y-values are 0 and -256/27, resulting in points (0,0) and (8/3,-256/27).

Step-by-step explanation:

To find the points on the graph of f(x) = x^3 - 4x^2 where the tangent line is horizontal, we look for where the derivative f'(x) is equal to zero, as a zero derivative indicates a slope of zero, which is the property of a horizontal line. We calculate the derivative of f(x) as follows:

1. Take the derivative of f(x) with respect to x: f'(x) = 3x^2 - 8x.
2. Set the derivative equal to zero and solve for x: 0 = 3x^2 - 8x, which simplifies to x(3x - 8) = 0.
3. We find the x-values where the derivative is zero: x = 0 and x = 8/3.
4. Now, find the corresponding y-values by plugging these x-values back into the original function: f(0) = 0^3 - 4*0^2 = 0 and f(8/3) = (8/3)^3 - 4*(8/3)^2.
5. Simplify to get the y-values: f(0) = 0 and f(8/3) = (512/27) - (256/9), which simplifies to (512/27) - (768/27), yielding -256/27.

Therefore, the points on the graph where the tangent line is horizontal are (0, 0) with the smaller x-value and (8/3, -256/27) with the larger x-value.