How do you find the value of c guaranteed by the mean value theorem if it can be applied for #f(x) = x^2 + 4x + 2# on the interval [-3,-2]?

Question

How do you find the value of c guaranteed by the mean value theorem if it can be applied for #f(x)  =  x^2 + 4x + 2# on the interval  [-3,-2]?

Charles Anctil · Accepted Answer

Given #f(x) = x^2+4x+2# The mean value theorem says that 
there exists a value #c epsilon [-3,-2]#
such that
#f'(c) = (f(-3) - f(-2))/((-3)-(-2)) = -1# #f'(x) = 2x+4# so
#f'(c)= 2c+4 = -1# #c = -2 1/2#
which is in the range #[-3,-2]#

William Abdalla · Answer

undefined To find the value of $ c $ guaranteed by the Mean Value Theorem for the function $ f(x) = x^2 + 4x + 2 $ on the interval $[-3, -2]$, follow these steps:

1. Calculate the average rate of change of $ f(x) $ over the interval $[-3, -2]$ using the formula:
$$ \text{Average rate of change} = \frac{f(-2) - f(-3)}{-2 - (-3)} $$

2. Evaluate $ f(-2) $ and $ f(-3) $ by substituting $ x = -2 $ and $ x = -3 $ into the function $ f(x) = x^2 + 4x + 2 $.

3. Once you have the average rate of change, set it equal to the derivative of $ f(x) $ with respect to $ x $ evaluated at some point $ c $ within the interval $[-3, -2]$:
$$ f'(c) = \text{Average rate of change} $$

4. Find the derivative $ f'(x) $ of $ f(x) $, then solve the equation $ f'(c) = \text{Average rate of change} $ for $ c $.

5. Once you find $ c $, verify that it lies within the interval $[-3, -2]$.

6. The value of $ c $ thus obtained is guaranteed by the Mean Value Theorem.