Question 1162542
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The zeros of the function *[tex \Large y\ =\ 2^x\ -\ x^2] tell you where *[tex \Large 2^x\ =\ x^2]. The intervals where *[tex \Large y\ >\ 0] give you the intervals where *[tex \Large 2^x\ >\ x^2].


Find the three zeros by inspection (-0.8 is close enough for the irrational one) which gives you four intervals.  Pick a value that is not close to the endpoint of each interval (choose the smallest integer in the interval) and calculate the value of the function.  If you get a positive result, then that interval is where *[tex \Large 2^x\ >\ x^2].  Or just do this part by inspection as well -- if the graph is above the *[tex \Large x]-axis, that interval is part of the solution set of the inequality.  You should end up with the union of two disjoint intervals.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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