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Arrange the functions in ascending order, starting with the function that eventually has the least value and ending with the function that eventually has the greatest value.

3x + 8

4x + 3

x2 + 6

3x

4x

x2


↓


↓


↓


↓


↓

Sagot :

Oladayoademosuidn

Answer:

3x < 4x < 3x + 8  < 4x + 3 < x² < x² + 6

Step-by-step explanation:

This is because as x → ∞

3x → 3∞ ,

Also, as x → ∞

4x → 4∞ ,

Also, as x → ∞

3x + 8  → 3∞ + 8,

Also, as x → ∞

4x + 3 → 4∞ + 3 ,

Also, as x → ∞

x² → ∞²,

Also, as x → ∞

x² + 6 → ∞² + 6

and 3∞ < 4∞ < 3∞ + 8 < 4∞  + 3 < ∞² < ∞² + 6

Thus as x → ∞

3x < 4x < 3x + 8  < 4x + 3 < x² < x² + 6 in ascending order

Madisonpuchalskiidn

Answer:

i am not 100% sure but i think it is

3x < 3x + 8 < 4^x < 4^x + 3 < x^2 < x^2 + 6

Step-by-step explanation:

I just got this because one of the other answers said replace x with infinite.  Their answer would have been wrong because the actual question is supposed to be 4^x and 4^x + 3 which is different than just 4x and 4x + 3. :)

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