V(t) was never zero on the interval [ 0, ln 2] according to the mean-value theorem.
According to the Mean Value Theorem, if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then f'(c) must equal the function's average rate of change over [a, b] at some point c on the interval (a, b).
given
Assume that f: [a, b] R and that f has a local maximum or minimum at x0 (a, b). F 0 (x0) = 0 if f is differentiable at x0.
Proof: Let's assume that f has a local maximum at x0 (a, b). When h is small enough, f(x0 + h) f. (x0).
f(x0 + h) f(x0) h 0 if h > 0 else.
Similar to this, f(x0 + h) f(x0) h 0 if h 0.
As a result of fundamental limit qualities, f
We point out that if x0 is either an or b, the prior theorem is invalid. For instance, f has a maximum at 1 but f 0 (x) = 1 for all x [0, 1] if we take the function f: [0, 1] R such that f(x) = x.
V(t) was never zero on the interval [0, ln 2] according to the mean-value theorem.
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