The function defined below satisfies Rolle's Theorem on the given interval. Find the value of c in the interval (0,1) where f'(c)=0.
f(x) = x3 - 2x2 + x, [0, 1]
Round your answer to two decimal places.
f(x) = x^3 - 2x^2 + x
f'(x) = 3x^2 - 4x + 1
so, where is f' = 0?
To find the value of c in the interval (0,1) where f'(c)=0 for the given function f(x) = x^3 - 2x^2 + x on the interval [0,1], we need to first find the derivative of the function.
To find the derivative, f'(x), of the function f(x), we need to take the derivative of each term. The derivative of x^n, where n is a constant, is nx^(n-1). Applying this rule to each term in the function:
f'(x) = d/dx(x^3) - d/dx(2x^2) + d/dx(x)
Simplifying each term:
f'(x) = 3x^2 - 4x + 1
To find the value of c in the interval (0,1) where f'(c) = 0, we need to set f'(x) equal to 0 and solve for x:
0 = 3x^2 - 4x + 1
This is a quadratic equation, so we can solve it by factoring, completing the square, or using the quadratic formula.
Let's use factoring to solve the equation:
0 = (3x - 1)(x - 1)
Setting each factor equal to zero:
3x - 1 = 0 or x - 1 = 0
Solving each equation:
3x = 1 or x = 1
x = 1/3 or x = 1
Since we are looking for a value of c in the interval (0,1), we can conclude that c = 1/3 is the value of x where f'(c) = 0 in the interval (0,1).