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Question -

Find the approximate value of┬аf┬а(5.001),where┬аf┬а(x) =┬аx3┬атИТ7x2┬а+ 15.



Answer -

Let┬аx┬а= 5 and ╬Фx┬а=0.001. Then, we have:

Hence, the approximate value of┬аf┬а(5.001)is тИТ34.995.

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