Find the coordinates of points where the graph of f(x) has horizontal tangents. As a check, graph f(x) and see whether the points you found look as though they have horizontal tangents.
f(x) = −x3 + 6x2 − 9x + 13
(x, y) = ( ? ) (smaller x-value)
(x, y) = ( ? ) (larger x-value)
To find the coordinates of points where the graph of f(x) has horizontal tangents, we need to find the values of x where the derivative of f(x) is equal to zero.
1. First, let's find the derivative of f(x):
f'(x) = d/dx(-x^3 + 6x^2 - 9x + 13)
To find the derivative, we can apply the power rule and differentiate each term separately:
f'(x) = -3x^2 + 12x - 9
2. Now, we set f'(x) equal to zero and solve for x:
-3x^2 + 12x - 9 = 0
To solve this quadratic equation, we can factor it:
-3(x^2 - 4x + 3) = 0
-3(x - 1)(x - 3) = 0
Setting each factor equal to zero gives us two x-values:
x - 1 = 0 --> x = 1
x - 3 = 0 --> x = 3
Therefore, the x-values where the graph of f(x) has horizontal tangents are x = 1 and x = 3.
3. To find the corresponding y-values, we substitute the x-values back into the original equation f(x):
For x = 1:
f(1) = -(1)^3 + 6(1)^2 - 9(1) + 13
f(1) = -1 + 6 - 9 + 13
f(1) = 9
For x = 3:
f(3) = -(3)^3 + 6(3)^2 - 9(3) + 13
f(3) = -27 + 54 - 27 + 13
f(3) = 13
So, the coordinates of the points where the graph of f(x) has horizontal tangents are (1, 9) and (3, 13).
Finally, graphing the function f(x) = −x^3 + 6x^2 − 9x + 13 and identifying (1,9) and (3, 13) will help visually verify that they indeed correspond to points where the tangent lines are horizontal.