# How do you find the absolute maximum and absolute minimum values of f on the given interval: #f(t) =t sqrt(25-t^2)# on [-1, 5]?

Reqd. extreme values are

Note that this substitution is allowed due to

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Find the monotony of the function from the derivative's sign and decide which local maximum/minimums are the biggest, smallest.

Absolute maximum is:

Absolute minimum is:

The function's derivative:

Regarding the highest possible value:

plot{xsqrt(25-x^2) [-14.4, 21.63, -5.14, 12.87]}

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To find the absolute maximum and absolute minimum values of ( f(t) = t \sqrt{25 - t^2} ) on the interval ([-1, 5]), you follow these steps:

- Find the critical points of ( f(t) ) within the interval by setting its derivative equal to zero and solving for ( t ).
- Evaluate ( f(t) ) at the critical points obtained in step 1, and also at the endpoints of the interval.
- Compare the values obtained in step 2 to determine the absolute maximum and absolute minimum values of ( f(t) ) on the interval.

The critical points of ( f(t) ) are found by solving the equation ( f'(t) = 0 ), where ( f'(t) ) is the derivative of ( f(t) ) with respect to ( t ). The derivative of ( f(t) ) is calculated using the product rule.

After finding the critical points and evaluating ( f(t) ) at those points and the endpoints of the interval, you compare the values to determine which one is the absolute maximum and which one is the absolute minimum.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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