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.
MVT --> Median Value Theorem
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
ohh that makes sense, thank you so much!! sorry I saw this so late
To apply the Mean Value Theorem (MVT) to the interval [1, x], we need to check three conditions:
1. f(x) must be continuous on the interval [1, x],
2. f(x) must be differentiable on (1, x), and
3. The slope of the secant line connecting the endpoints of the interval must be equal to the derivative at some point in the interval.
Let's verify each condition.
1. Continuity:
We are given that f(x) is differentiable for x ≥ 0, so it is also continuous on this interval. Since x > 1, it falls within the domain of f(x). Therefore, f(x) is continuous on the interval [1, x].
2. Differentiability:
We are given that f'(x) ≤ 2 for x ≥ 0. Since x > 1, it is within the domain of f'(x). Thus, f(x) is differentiable on the interval (1, x).
3. Equality of Secant Slope and Derivative:
To apply the MVT, we need to show that the slope of the secant line connecting (1, f(1)) and (x, f(x)) is equal to f'(c) for some c in the interval (1, x).
The slope of the secant line is given by (f(x) - f(1)) / (x - 1). Substituting f(1) = 1, we have:
slope = (f(x) - 1) / (x - 1)
By the MVT, there exists a c in (1, x) such that f'(c) = slope. Since we are given f'(x) ≤ 2 for x ≥ 0, we have:
f'(c) ≤ 2
Substituting the slope, we get:
(f(x) - 1) / (x - 1) ≤ 2
Rearranging and simplifying, we have:
f(x) - 1 ≤ 2(x - 1)
f(x) ≤ 2x - 2 + 1
f(x) ≤ 2x - 1
Therefore, we have proven that for all x > 1, f(x) ≤ 2x - 1 using the Mean Value Theorem, given the conditions f(1) = 1 and f'(x) ≤ 2 for x ≥ 0.