Explanation

Explanation

Explanation

Explanation

Let the quadratic polynomial be f(x) = ax^2 + bx + c.

Given that f'(1) = 1 and f''(1) = 2.

f'(x) = 2ax + b

f''(x) = 2a

f'(1) = 1 => 2a + b = 1

f''(1) = 2 => 2a = 2 => a = 1

Substituting a = 1 in the first equation, we get:

2 + b = 1 => b = -1

Now, f(x) = x^2 - x + c

f(1) = 1^2 - 1 + c = c

f(0) = 0^2 - 0 + c = c

f(1) - f(0) = c - c = 0

Explanation

• 1. The functions f(x) and g(x) are continuous on [a, b].
• 2. The functions f(x) and g(x) are differentiable on (a, b).
• 3. g'(x) ≠ 0 for any x in (a, b).

Explanation

Explanation

Explanation

Explanation

To find the value of c that satisfies the Mean Value Theorem, we need to find the derivative of the function f(x) = x^2 + 2x + 1.

f'(x) = 2x + 2

The Mean Value Theorem states that there exists a point c in the interval [1, 2] such that:

f'(c) = (f(2) - f(1)) / (2 - 1)

First, let's calculate f(2) and f(1):

f(2) = 2^2 + 2(2) + 1 = 9

f(1) = 1^2 + 2(1) + 1 = 4

Now, we can calculate the slope:

(f(2) - f(1)) / (2 - 1) = (9 - 4) / 1 = 5

Now, we set f'(c) equal to 5:

2c + 2 = 5

Solving for c, we get:

2c = 3

c = 3/2

Since c = 3/2 is between 1 and 2, it satisfies the Mean Value Theorem.

Explanation

To form a differential equation from a given function containing arbitrary constants:

• If the function has 'n' arbitrary constants, differentiate the equation 'n' times with respect to the independent variable.
• This process removes the arbitrary constants, leaving a differential equation.