Asked by maath
Suppose f(1)=1 and f'(x)≤2 for x≥0. Apply the MVT to the interval [1,x] to prove that f(x)≤2x-1 for all x>1. Verify all the conditions of MVT are satisfied.
Answers
Answered by
maath
MVT --> Median Value Theorem
Answered by
oobleck
on the interval [1,x] the MVT states that there is a number c such that
(f(c) - f(1))/(c-1) = f'(c)
Now, we know that f'(x) <= 2 on [1,x], so
(f(c) - f(1))/(c-1) <= 2
Now, f(c) - f(1) = f(c) - 1
so, f(c) - 1 <= 2(c-1)
f(c) <= 2c-1
(f(c) - f(1))/(c-1) = f'(c)
Now, we know that f'(x) <= 2 on [1,x], so
(f(c) - f(1))/(c-1) <= 2
Now, f(c) - f(1) = f(c) - 1
so, f(c) - 1 <= 2(c-1)
f(c) <= 2c-1
Answered by
maath
ohh that makes sense, thank you so much!! sorry I saw this so late
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