Suppose that f(t)f(t) is continuous and twice-differentiable for t≥0t≥0. further suppose f′′(t)≤3f″(t)≤3 for all t≥0t≥0 and f(0)=f′(0)=0f(0)=f′(0)=0. using the racetrack principle, what linear function g(t)g(t) can we prove is greater than or equal to f′(t)f′(t) (for t≥0t≥0)?

(x) is continuous and twice differentiable for ±≥ 0. Suppose f”( ±) ≤ 5 for all ±≥ 0 and f (0) = f’(0) = 0. The linear function with a derivation of 0 is y = 0. Thus we have g(t) = 0. The quadratic function with concavity 5 and initial slope 0 is y = 5/2 x^2 + 0x. Therefore, h(t) = (5/2)x^2

gmt is 4 hrs ahead of edt so just add four hours to 1: 14: 20 (minutes and seconds stay the same)

gmt is 5: 14: 20

hope this !

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