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Range of the f(x)(e^x - 1)/(e^x+ 1) |
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Answer» ong>ANSWER: (-1, 1) Step-by-step EXPLANATION: for x = 0, numerator becomes 0 1) for a any positive value of n numerator is 2 less than DENOMINATOR, therefore fraction is ALWAYS lesser than 1. 2) for a NEGATIVE value of n numerator becomes (1/eⁿ -1) which is a negative value but denominator remains positive, so values of f(x) is negative, for a greater values of x like -∞ values will be closer to -1. therefore range is (-1, 1) |
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