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Apply the Mean Value Theorem and find a value of c in the interval
[4,9] that satisfies the equation [tex]f(x)=\sqrt{x}[/tex]

Respuesta :

musiclover10045

If f(x) is continuous on [a,b] and differtiable on the open interval (a,b) then c exists such that f’(c) = f(b) - f(a) / b-a.

Using the given function f(x) = sqrt (x)

And the given interval [4,9]

F(c) = sqrt(9) - sqrt (4) / 9-4

= 3-2 / 5

= 1/5

C = 1/5

amna04352

Answer:

c = 1444/225

Step-by-step explanation:

1/(9-4) × integral sqrt(x)

1/5 [⅔ × x^(3/2)]

Apply limits

(2/15)[9^(3/2) - 4^(3/2)]

(2/15)(27 - 8)

(2/15)(19)

38/15

38/15 = sqrt(c)

c = 1444/225

Or, 6.41777777778

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