Mean Value Theorem range
f ( x ) is a differentiable function that satisfies 5 ≤ f ′ ( x ) ≤ 1 4 for all x . Let a and b be the maximum and minimum values, respectively, that f ( 1 1 ) − f ( 3 ) can possibly have, then what is the value of a + b ?
The answer is 152.
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2 solutions
f ( x ) is a differentiable function that satisfies 5 ≤ f ′ ( x ) ≤ 1 4 for all x . Let a and b be the maximum and minimum values, respectively, that f ( 1 1 ) − f ( 3 ) can possibly have, then what is the value of a + b ?
Since 5 ≤ f ′ ( x ) ≤ 1 4 , then 5 ≤ 1 1 − 3 f ( 1 1 ) − f ( 3 ) ≤ 1 4 . Simplifying this gives 4 0 ≤ f ( 1 1 ) − f ( 3 ) ≤ 1 1 2 . Thus a = 1 1 2 , b = 4 0 ⇒ a + b = 1 5 2 .
Since f ( x ) is differentiable function, then by MVT ∃ c ∈ ( 3 , 1 1 ) ∋ 1 1 − 3 f ( 1 1 ) − f ( 3 ) = f ′ ( c ) ⇒ f ( 1 1 ) − f ( 3 ) = 8 f ′ ( c )
Because 5 ≤ f ′ ( x ) ≤ 1 4 , so we have 8 ( 5 ) ≤ f ( 1 1 ) − f ( 3 ) ≤ 8 ( 1 4 ) ⇒ a + b = 4 0 + 1 1 2 = 1 5 2