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Andrew [12]
2 years ago
13

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)?
Mathematics
1 answer:
Ivan2 years ago
4 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

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Answer:

  x > 4/3

Step-by-step explanation:

The first inequality can be solved this way ...

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  3x > 4 . . . . . . . add 91

  x > 4/3 . . . . . . divide by 3

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The second inequality has solution ...

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The solution set is the union of these overlapping solutions, so will be equal to the first solution:

  x > 4/3

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Part 1) Choose the two equations that represent the situation

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<em>Non-member</em>

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