Tyler York

June 9, 2023

In this video, Achievable GRE course author Matt Roy explains how to approach problems that involve a function inside another function. The key being, to solve for the output of the interior function first, and only then solving for the output of the exterior function.

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00:00:02 This problem is a functions problem with embedded functions, and the key to any problem like this is to solve for the interior function before solving for the exterior function. That is, you find the input for the interior function, solve for the output, and that output becomes the input for the exterior function. So let's start off with quantity A, quantity A. 00:00:26 Is the F of the G of four, so we should start off with the G of four. So inputting 4 for X in the G function, we get 2 / 4 + 3, which is 2 / 7. Now 2 / 7 becomes the output of the interior function, which we will use as the input of the exterior function. Now we input 2 / 7 into the F function 2 / 7 + 3. 00:00:56 Over 2 becomes 23 / 7 / 2 because we we should just add with a least common multiple as the base of the of both or as the denominator of both fractions. So it becomes 2 / 7 + 21 / 7 because 2121 / 7 is 3. So dividing by two is the same as multiplying by its reciprocal, which is 1 / 2. 00:01:24 So now we have 23 / 7 * 1 / 2. This is equivalent to 23 / 14 for quantity B. We have the G of the F of four. So the functions are flipped, but with the same interior function input. We should start off with the F of four. The F of 4 = 4 + 3 / 2, which is 7 / 2. Seven over 2, which is our output of the interior function, becomes. 00:01:53 The input of the exterior function, which is the G function. So 7 / 2 becomes the input of the G function, meaning it becomes 2 / 7 / 2 + 3. Now 7 / 2 + 3 is equivalent to 7 / 2 + 6 / 2, which would give us 13 / 2, so we have 2 / 13 / 2. Again, dividing by a fraction is the same as multiplying by its reciprocal. 00:02:23 So we end up with 2 * 2 / 13, which is equivalent to 4 / 13. Now obviously that's a fraction less than one, and quantity A is greater than one, so quantity A is greater.

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