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) = x^3 - 2x^2 + x, [0, 1]
Round your answer to two decimal places.
To find the value of c where f'(c) = 0, we need to follow these steps:
1. Find the derivative of the function f(x).
2. Set the derivative equal to zero and solve for x to find the critical points where f'(x) = 0.
3. Check if the critical points are in the interval (0, 1).
4. If there is a critical point in the interval (0, 1), that value of x will be equal to c.
Let's follow these steps to find the value of c:
Step 1: Find the derivative of f(x)
To find the derivative of f(x), we differentiate each term of the function using the power rule. The derivative of x^n is n * x^(n-1).
f(x) = x^3 - 2x^2 + x
f'(x) = (3x^2) - (4x) + (1)
Step 2: Set f'(x) = 0 and solve for x
Now we set f'(x) = 0 and solve for x to find the critical points.
(3x^2) - (4x) + (1) = 0
Since this is a quadratic equation, we can solve it by factoring, completing the square, or using the quadratic formula.
By factoring, we find:
(3x - 1)(x - 1) = 0
This gives two possible critical points:
3x - 1 = 0 => x = 1/3
x - 1 = 0 => x = 1
Step 3: Check if the critical points are in the interval (0, 1)
Now we need to check if these critical points, x = 1/3 and x = 1, fall within the interval (0, 1).
Since both 1/3 and 1 are within the interval (0, 1), they are valid values for c.
Step 4: Choose the appropriate value of c
We are asked to round our answer to two decimal places. Since 1/3 has infinitely repeating decimals, we should choose 1 as the value of c.
Therefore, the value of c in the interval (0, 1) where f'(c) = 0 is c = 1.
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