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Doss [256]
3 years ago
15

1. Verify that the function

Mathematics
1 answer:
Novay_Z [31]3 years ago
6 0

Answer:

0 and 8/3

Step-by-step explanation:

y is continuous on [0,4]

It also also differentiable on (0,4)

Plugging in 0 and 4, we get -3 both times so y(0) = y(4).

Therefore there is at least one value c between 0 and 4 inclusive where y'(c) = 0

y' = -3x^2 + 8x

Set that to 0:

-3x^2 + 8x = 0

x(-3x+8) = 0

x = 0 or 8/3

And both values work.

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A) You can have

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