The function f is continuous on the closed interval closed interval from pi over 3 to 5 times pi all over 3.. Selected values of f are given in the table: x pi over 3 | 2 times pi all over 3 | 4 times pi all over 3 | 5 times pi all over 3
f(x) 0.4775 | −0.2387 | −0.1194. | k

What value of k in the table would give the equation f(x) = 0 at least two solutions in the interval closed interval from pi over 3 to 5 times pi all over 3?

negative pi over 3
−0.0955
0.0955
negative 5 times pi all over 3

step-by-step explanation:

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step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)