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.
First, be sure that f(0)=f(1). It does
since f'(x) = 3x^2-4x+1 = (3x-1)(x-1)
just find a value in the interval [0,1] where f'(x) = 0. Now it's just algebra. The calculus is done.
To find the value of c in the interval (0,1) where f'(c)=0, follow these steps:
Step 1: Calculate the derivative of the function f(x). Let's call it f'(x).
f'(x) = 3x^2 - 4x + 1
Step 2: Set f'(x) equal to 0 and solve for x.
3x^2 - 4x + 1 = 0
Step 3: Use the quadratic formula to solve for x.
The quadratic formula states that for ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 3, b = -4, and c = 1. Plugging these values into the quadratic formula gives:
x = (4 ± √((-4)^2 - 4*3*1))/(2*3)
x = (4 ± √(16 - 12))/(6)
x = (4 ± √(4))/(6)
x = (4 ± 2)/(6)
Simplifying, we get two possible solutions:
x = (4 + 2)/(6) = 6/6 = 1
x = (4 - 2)/(6) = 2/6 = 1/3
Step 4: Check if the solutions are within the interval (0,1).
Both solutions, x = 1 and x = 1/3, are within the interval (0,1).
Step 5: Round the answer to two decimal places.
The value of c in the interval (0,1) where f'(c) = 0 is approximately 1.00 when rounded to two decimal places.