Question

X
Let f(a) = [ + (6 – 3+ + 2) de
1
1 < x < 4
(A) The global maximum of f (x)
in [1, 4) is 0
(B) The global maximum of f (x
)
is
63
4
(C) The global maximum of f (x)
in 1 < x < 4is -
1
4
(D) The global maximum of f (x
in 1 < x < 4is 2.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

f(x)=log

10

(4x

3

−12x

2

+11x−3)

xϵ[2,3]

f(2)=log

10

(32−38+22−3)

f(2)=log

10

3

f(3)=log

10

(108−108+11×3−3)

=log

10

(30)

=log

10

3+log

10

10

f(3)=1+log

10

3

Therefore, f(x) will have maximum value of 1+log

10

3