How do you estimate the instantaneous rate of change of #f(t) =sqrt(9t + 6)# for #t = 5#?

Answer 1

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Answer 2

To estimate the instantaneous rate of change of ( f(t) = \sqrt{9t + 6} ) for ( t = 5 ), you can use the derivative of the function. First, find the derivative of ( f(t) ) with respect to ( t ), then evaluate it at ( t = 5 ).

The derivative of ( f(t) = \sqrt{9t + 6} ) with respect to ( t ) is given by: [ f'(t) = \frac{d}{dt}(\sqrt{9t + 6}) = \frac{1}{2\sqrt{9t + 6}} \cdot \frac{d}{dt}(9t + 6) = \frac{9}{2\sqrt{9t + 6}} ]

Now, evaluate ( f'(t) ) at ( t = 5 ): [ f'(5) = \frac{9}{2\sqrt{9(5) + 6}} = \frac{9}{2\sqrt{45 + 6}} = \frac{9}{2\sqrt{51}} ]

So, the estimated instantaneous rate of change of ( f(t) = \sqrt{9t + 6} ) at ( t = 5 ) is ( \frac{9}{2\sqrt{51}} ).

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Answer from HIX Tutor

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.

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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