`f(t) = tsqrt(4 - t^2), [-1, 2]` Find the absolute maximum and minimum values of f on the given interval

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Function has absolute minimum=- `sqrt(3)` at t=-1 

Given: `f(t)=tsqrt(4-t^2),[-1,2].`

Find the critical number by setting the first derivative equal to zero and solving for the t values. Find the derivative using the product rule.

`f'(t)=t[-t/(4-t^2)^(1/2)]+(4-t^2)^(1/2)=0`

`t^2/(4-t^2)^(1/2)=(4-t^2)^(1/2)`

`t^2=4-t^2`

`2t^2=4`

`t^2=2`

`t=+-sqrt(2)` 

The critical numbers are `x=+-sqrt(2).`

Substitute the critical numbers and the endpoints of the interval [-1,2] into the original f(t) function. Do NOT substitute in the `t=-sqrt(2)`  because it is not...

(The entire section contains 3 answers and 284 words.)

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