Question

For the given function, find all values of c over the interval (1, 4) stipulated by Rolle's Theorem such that f'(c) = 0. f(x) = x² / (5x-4)

Answer 1

The only value of c within the interval (1, 4) such that f'(c) = 0 is c = 8/5.

Step-by-step explanation:

To find the values of c over the interval (1, 4) stipulated by Rolle's Theorem such that f'(c) = 0, we need to find the derivative of the function f(x) = x² / (5x-4) and solve for x when f'(x) = 0. First, let's find the derivative:

f'(x) = (2x(5x - 4) - x²(5)) / (5x - 4)²

f'(x) = (10x² - 8x - 5x²) / (5x - 4)²

f'(x) = (5x² - 8x) / (5x - 4)²

Now, let's solve for x when f'(x) = 0:

(5x² - 8x) / (5x - 4)² = 0

5x² - 8x = 0

x(5x - 8) = 0

x = 0 or 5x - 8 = 0

x = 0 or x = 8/5

However, we need to find the values of c, which are within the interval (1, 4). So, x = 0 is not within the interval. Therefore, the only value of c within the interval (1, 4) such that f'(c) = 0 is c = 8/5.

Answer 2

There are no values of c that satisfy Rolle's Theorem for the given function over the interval (1, 4).

Step-by-step explanation:

To find all values of c over the interval (1, 4) stipulated by Rolle's Theorem such that f'(c) = 0 for the given function f(x) = x² / (5x-4), we need to find the derivative of the function and set it equal to 0.

The derivative of f(x) can be found using the quotient rule:

f'(x) = (d/dx)(x²) / (5x-4) - x²(d/dx)(5x-4) / (5x-4)²

Simplifying this expression, we get:

f'(x) = (2x(5x-4) - x²(5)) / (5x-4)²

Setting f'(x) equal to 0, we have:

(2x(5x-4) - x²(5)) / (5x-4)² = 0

Solving this equation for x, we find:

10x² - 8x - 5x² = 0

-3x² - 8x = 0

-x(3x + 8) = 0

So, the values of x that make f'(x) = 0 are x = 0 and x = -8/3.

However, we need to find the values of c over the interval (1, 4). Since neither of these values fall within this interval, there are no values of c that satisfy the conditions of Rolle's Theorem for this function over the given interval.