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

Sagot :

Answer:

Explanations:

According to the Mean Value Theorem:

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

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

[tex]\begin{gathered} 2c\text{ - 6 = }\frac{24-8}{8} \\ 2c\text{ - 6 = }\frac{16}{8} \\ 2c\text{ - 6 = 2} \\ 2c\text{ = 2 + 6} \\ 2c\text{ = 8} \\ c\text{ = }\frac{8}{2} \\ c\text{ = 4} \end{gathered}[/tex]

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