The domain of defination of the function f(x)=√22x+64(x-2)/3-2-1(72+22x)
i m not givin the options coz u'll be able to eliminate it(option) easily.solve wid proper method.
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2 Answers
24just apply all the respective domains togetherand take intersection
22x+64(x-2)/3-2-1(72+22x)≥0
& 22x>0
& 64(x-2)/3 >0
341You need the expression under the square root sign to be non-negative 2^{2x} + 64^{\frac{x-2}{3}} - \frac{1}{2} \left(72 + 2^{2x} \right) = 2^{2x} + \frac{2^{2x}}{16} - \frac{2^{2x}}{2} -36 \\ \\ = \frac{9}{16} 2^{2x} - 36 \ge 0 \Rightarrow 2^{2x} \ge 64 \Rightarrow 2x \ge 8 \\ \\ \Rightarrow \boxed{x \ge 4}
edit: should be 2x≥6, so x≥3
