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3. Use the intermediate value theorem to show that the equation has a solution on the given interval. (3)1132 5 =0 on (2,3); (b) 3:3

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3. Use the intermediate value theorem to show that the equation has a solution on the given interval. (3)1132 5 =0 on (2,3); (b) 3:3 33:2+ 1 = 0 on (0,1)

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