Let f(x) = x3 − 4x2.
Find the point(s) on the graph of f where the tangent line is horizontal.
f' = 3x^2 - 8x
so, where is f' = 0?
To find the point(s) on the graph of f(x) where the tangent line is horizontal, we need to first take the derivative of the function f(x) and find the critical points.
1. Find the derivative of f(x):
f'(x) = (d/dx) (x^3 - 4x^2) = 3x^2 - 8x
2. Set the derivative equal to 0 to find critical points:
3x^2 - 8x = 0
Factor out x: x(3x - 8) = 0
Therefore, x = 0 or x = 8/3 are the critical points.
3. Now we need to find the corresponding y-values for these critical points.
f(0) = (0)^3 - 4(0)^2 = 0
f(8/3) = (8/3)^3 - 4(8/3)^2
Simplify the expression.
And there you have it! By using the derivative and critical point analysis, we can find the x-values where the tangent line of f(x) is horizontal, and we can also find the corresponding y-values.