The function f(x)=x(x-1)(x-2) has stationary points at x=0, x=1 and x=2.true or false.
To determine if the function f(x) = x(x-1)(x-2) has stationary points at x=0, x=1, and x=2, we need to find the points where the derivative of the function equals zero.
Step 1: Find the derivative of f(x):
f'(x) = d/dx [ x(x-1)(x-2) ]
= (x-1)(x-2) + x(x-2) + x(x-1) (using the product rule)
= (x^2 - 3x + 2) + (x^2 - 2x) + (x^2 - x)
= 3x^2 - 6x + 2 - 2x^2 + 2x - x^2
= 0
Step 2: Set f'(x) equal to zero and solve for x:
3x^2 - 6x + 2 - 2x^2 + 2x - x^2 = 0
-2x^2 - 4x + 2 = 0
2(x^2 + 2x - 1) = 0
Step 3: Solve the quadratic equation for x:
x^2 + 2x - 1 = 0
Using the quadratic formula, we can find the solutions for x:
x = (-2 ± √(2^2 - 4*1*(-1))) / (2*1)
x = (-2 ± √(4 + 4)) / 2
x = (-2 ± √8) / 2
x = (-2 ± 2√2) / 2
x = -1 ± √2
Hence, the solutions for x are x = -1 + √2 and x = -1 - √2.
Since x = 0, x = 1, and x = 2 are not solutions of f'(x) = 0, we can conclude that the function f(x) = x(x-1)(x-2) does not have stationary points at x = 0, x = 1, and x = 2.
Therefore, the statement "The function f(x) = x(x-1)(x-2) has stationary points at x = 0, x = 1, and x = 2" is false.