The function g(x) is continuous on the closed interval [8, 10] and differentiable on the open interval
(8, 10). The value x = 8.5 satisfies the conditions of the Mean Value Theorem on the interval [8, 10].
What is g(10) given g'(8.5) = -9 and g(8) = 6?

Answer:g(10) = -12Step-by-step explanation:Mean Value TheoremIf f is continuous on [a, b] and differentiable on (a, b), then there is a number c  such that a < c < b and:[tex]f'(c)=\dfrac{f(b)-f(a)}{b-a}[/tex]We are told that the function g(x) is continuous on the closed interval [8, 10] and differentiable on the open interval (8, 10).Therefore:a = 8b = 10Given:g'(8.5) = -9g(8) = 6As 8 < 8.5 < 10 then c = 8.5:[tex]\begin{aligned} \implies g'(8.5)=\dfrac{g(10)-g(8)}{10-8}&=-9\\\\\dfrac{g(10)-6}{2}&=-9\\\\g(10)-6&=-18\\\\g(10)&=-12\end{aligned}[/tex]