f(x) = (6x-9)/(x-2)
The Intermediate Value Theorem is not applicable in this case because f(x) is not continuous on the interval [1, 5] (f(x) is undefined at x = 2).
Mark M.
tutor
f(x) = (6x - 9)/(x-2) is not continuous at x = 2 because the denominator is equal to zero when x = 2. The Intermediate Value Theorem is applicable only for continuous functions. Did I misinterpret the function??
If f(x) = 6x - 9/x - 2 (leaving out the parentheses), then the function is continuous on [1, 5], f(1) < 4 and f(5) > 4. So, by the Intermediate Value Theorem, there is a number, c, between 1 and 5 so that f(c) = 4.
Find c: 6c - 9/c - 2 = 4
6c2 - 9 -2c = 4c
6c2 -6c - 9 = 0
Divide by 3: 2c2 - 2c - 3 = 0
c = (2 ± √[(-2)2-4(2)(-3)])/4
= (2 ± √28)/4 ≈ 1.8, -0.8
The value of c between 1 and 5 such that f(c) = 4 is
(2 + √28)/4 = (2 + 2√7)/4
= (1 + √7)/2
Report
05/03/15
Kim S.
05/03/15