The Mean Value Theorem :- If f : [a,b] rarr RR is (1) continuous on
[a,b] and (2) differentiable on (a,b), then EE c in (a,b) such that,
f'(c)=(f(b)-f(a))/(b-a).
Here, f : [3,6] rarr RR : f(x)=-x^2+8x-17.
f is a Quadratic Poly. We know that Polys. are cont. &
differentiable on RR, hence f is cont. on [3,6] sub RR, &, it is
differentiable on (3,6).
:. f'(c)=(f(6)-f(3))/(6-3)," for some c in "[3,6].................(star).
Now, f(x)=-x^2+8x-17 :. f'(x)=-2x+8
:. f'(c)=-2c+8....................................................(1)
Also,
(f(6)-f(3))/(6-3)={(-36+48-17)-(-9+24-17)}/3, i.e.,
(f(6)-f(3))/(6-3)=-3/3=-1...............................(2).
Hence, by (1),(2), and, (star), we have,
-2c+8=-1 rArr -2c=-9 rArr c=9/2=4.5 in (3,6).
Enjoy maths.!