What is the end behavior of the function f(x)=2/3x^3? As x→[infinity], f(x)→−[infinity] As x→−[infinity], f(x)→−[infinity] As x→[infinity], f(x)→[infinity] As x→−[infinity], f(x)→[infinity] As x→[infinity], f(x)→[infinity] As x→−[infinity], f(x)→−[infinity] As x→[infinity], f(x)→−[infinity] As x→−[infinity], f(x)→[infinity]

As x→[infinity], f(x)→[infinity] As x→−[infinity], f(x)→−[infinity]

Step-by-step explanation:

Since f(x) =2x³/3,

Let x → + ∞

Substituting x = + ∞ into the equation, we have

f(+∞) = 2(+∞)³/3

= + 2∞³/3

= + ∞            (since + 2∞³/3 → +∞ (a larger number))

So as x → + ∞, f(x) → + ∞

Let x → - ∞

Substituting x = - ∞ into the equation, we have

f(-∞) = 2(-∞)³/3

= - 2∞³/3

= - ∞         (since - 2∞³/3 → -∞ (a larger number))

So as x → -∞, f(x) → -∞

So, the answer is As x→[infinity], f(x)→[infinity] As x→−[infinity], f(x)→−[infinity]

p> q if you calculate p is100 q is 79

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step-by-step explanation: