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.!