Verified answer

Apply the Mean Value Theorem to f on [3, 6]:

[f(6) - f(3)] / (6 - 3) = f '(c) for some c in (3, 6).

Since f '(x) ≥ 2 for 3 ≤ x ≤ 6,

we can rewrite this as [f(6) - f(3)] / (6 - 3) ≥ 2.

Thus, [f(6) - 7] / 3 ≥ 2 ==> f(6) ≥ 13.

So, the smallest value for f(6) is 13.

I hope this helps!

... ƒ'(x) ≥ 2 ... for ... 3 ≤ x ≤ 6

∴ ∫ ƒ'(x) dx ≥ ∫ 2 dx ... on [3, 6]

∴ [ ƒ(x) ] ≥ [ 2x ] .......... on [3. 6]

∴ ƒ(6) - ƒ(3) ≥ [ 2(6) - 2(3) ]

∴ ƒ(6) - 7 ≥ 6 ................................ [ given: ƒ(3) = 7 ]

∴ ƒ(6) ≥ 13

∴ Minimum of ƒ(6) can be = 13. ................................... Ans.

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