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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?

Sagot :

Semsee45idn

Answer:

g(10) = -12

Step-by-step explanation:

Mean Value Theorem

If 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 = 8
  • b = 10

Given:

  • g'(8.5) = -9
  • g(8) = 6

As 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]

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