How do you find the number c that satisfies the conclusion of the Mean Value Theorem for the function #f(x)=x^2 - 2x + 5# on the interval #[1, 3]#?

Question

How do you find the number c that satisfies the conclusion of the Mean Value Theorem for the function #f(x)=x^2 - 2x + 5#  on the interval #[1, 3]#?

Elijah Abram · Accepted Answer

Solve #2x-2=2# One way to write the conclusion of the Mean Value Theorem for the function #f(x)# on the interval #[a, b]# is: There is a #c in (a,b)# such that #f'(c) = (f(b)-f(a))/(b-a)#. In this question, we have #f(x)=x^2 - 2x + 5# on the interval #[1, 3]# So, find #f'(x)#. Find #(f(3)-f(1))/(3-1)#. Set them equal to one another and solve the resulting equation. If the equation has more than 1 solution, list only the solutions in the interval #(1,3)#.

Axel Alvarez · Answer

undefined To find the number $ c $ that satisfies the conclusion of the Mean Value Theorem for the function $ f(x) = x^2 - 2x + 5 $ on the interval $[1, 3]$, follow these steps:

1. Find the derivative of $ f(x) $, denoted as $ f'(x) $.
2. Evaluate $ f'(x) $ at each endpoint of the interval $[1, 3]$ to find the values of $ f'(1) $ and $ f'(3) $.
3. Compute the average rate of change of $ f(x) $ over the interval $[1, 3]$ using the formula: 
   $$ \text{Average rate of change} = \frac{f(3) - f(1)}{3 - 1} $$
4. By the Mean Value Theorem, there exists a number $ c $ in the interval $[1, 3]$ such that $ f'(c) $ equals the average rate of change computed in step 3.
5. Set $ f'(c) $ equal to the average rate of change and solve for $ c $.

Let's go through the steps:

1. $ f'(x) = 2x - 2 $
2. Evaluate:
   - $ f'(1) = 2(1) - 2 = 0 $
   - $ f'(3) = 2(3) - 2 = 4 $
3. Average rate of change:
   $$ \text{Average rate of change} = \frac{f(3) - f(1)}{3 - 1} = \frac{(3^2 - 2(3) + 5) - (1^2 - 2(1) + 5)}{3 - 1} = \frac{9 - 6 + 5 - 1 + 2 + 5}{2} = \frac{8}{2} = 4 $$
4. Since $ f'(1) = 0 $ and $ f'(3) = 4 $, the function $ f(x) $ is continuous and differentiable on the interval $[1, 3]$.
5. According to the Mean Value Theorem:
   $$ f'(c) = \frac{f(3) - f(1)}{3 - 1} = 4 $$
   Solve for $ c $:
   $$ 2c - 2 = 4 $$
   $$ 2c = 6 $$
   $$ c = 3 $$

Thus, $ c = 3 $ satisfies the conclusion of the Mean Value Theorem for the given function on the interval $[1, 3]$.