I'm quite confused with the racetrack principle and need on this problem. . suppose that f(t) is continuous and twice-differentiable for t> = 0. further suppose f''(t) > = 3 for all t> = 0 and f(0) = f'(0) = 0.. using the racetrack principle, what linear function g(t) can we prove is less than f'(t) (for t> = 0)? . g(t) = . then, also using the racetrack principle, what quadratic function h(t) can we prove is less than than f(t) (for t> = 0)? . h(t) = .

Race track principle says that if two functions are equal at t=0, then the one which has a greater derivative will be greater.

In this case we're comparing f′(t) and g′(t). So we make sure that g(0)=f′(0) and that f′′(t)≥g′(t)

g(t)=at+b

Since it is a line.
g′(t)=a

f′′(t)≥3≥g′(t)⟹3≥a

So let a=3.

So

h(t)=at2+bt

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

christopher is 272,195 year old

step-by-step explanation: