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MVT of a function x^2-6x+8 on (0,8)

1 Answer

5 votes
Answer:

Explanations:

According to the Mean Value Theorem:


f^(\prime)(c)\text{ = }(f(x_2)-f(x_1))/(x_2-x_1)
f^(\prime)(c)\text{ = }(f(8)-f(0))/(8-0)

f(x) = x² - 6x + 8

f(0) = 0² - 6(0) + 8

f(0) = 8

f(8) = 8² - 6(8) + 8

f(8) = 64 - 48 + 8

f(8) = 24

f'(x) = 2x - 6

f'(c) = 2c - 6


\begin{gathered} 2c\text{ - 6 = }(24-8)/(8) \\ 2c\text{ - 6 = }(16)/(8) \\ 2c\text{ - 6 = 2} \\ 2c\text{ = 2 + 6} \\ 2c\text{ = 8} \\ c\text{ = }(8)/(2) \\ c\text{ = 4} \end{gathered}

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