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Current time:0:00Total duration:4:02

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.4 (EK)

, FUN‑4.A.5 (EK)

, FUN‑4.A.6 (EK)

olga was asked to find where f of X is equal to X minus 2 to the fourth power has inflection points this is her solution so we look at her solution and then they ask us is Olga's work correct if not what's her mistake so pause this video and see if you can figure this out all right let's just just follow her work so here she's trying to take the first derivative so you would apply the chain rule it would be 4 times X minus 2 to the third power times the derivative of X minus 2 which is just 1 so this checks out then you take the derivative of this would be 3 times 4 which would be 12 times X minus 2 to the second power times the derivative of X minus 2 which is just 1 which is exactly what she has here 12 times X minus 2 to the second power that checks out so step ones looking good for olga step 2 the solution of the second derivative equaling zero is x equals 2 and that looks right the second derivative is 12 times X minus 2 squared and we want to make that equal to zero this is only going to be true when X is equal to two so step two is looking good so step three Olga says f has an inflection point at x equals two so she's basing this just on the fact that the second derivative is zero when X is equal to 2 shes basing this just on the fact that f prime prime of 2 is equal to 0 now I have a problem with this because the fact that your second derivative is 0 at x equals 2 that makes 2 a nice candidate to check out but you can't immediately say that we have an inflection point there remember an inflection point is where we go from being concave upwards to concave downwards or concave downwards to concave upwards and speaking in the language of the second derivative it means that the second derivative changes signs as we go from below x equals 2 to above X equals 2 but we have to test that because it's not necessarily always the case so let's actually test it let's think about some intervals intervals so let's think about the interval when we go from negative infinity to 2 and let's think about the interval where we go from 2 to positive infinity if you want you could have some test values could think about the sign sign of our second derivative and then based on that you could think about concavity concavity of F so let's think about what's happening so you could take a test value let's say one is in this interval and let's say three is in this interval and you could say one minus two squared is going to be let's see that's negative one squared which is one and then you're just going to this is just going to be twelve so this is going to be positive and if you tried three three minus two squared is one times 12 well that's also going to be positive and so you're going to be concave upwards at least at these test values it looks like on either side of two that the sign of the second derivative is positive on either side of two and you might say well maybe I just need to find closer values but if you inspect the second derivative here you can see that this is never going to be negative in fact for any value other than x equals two and this value right over here since we're even if X minus two is negative you're squaring it which will make this entire thing positive and then multiplying it times a positive value so for any value other than x equals two the sign of our second derivative is positive which means that we're going to be concave upwards and so we actually don't have an inflection point at x equals two because we are not switching signs as we go from values less than x equals two to values greater than x equals two our second derivative is not switching signs so once again this is incorrect we actually don't have an inflection point at x equals two because our second derivative does not switch signs as we cross x equals two which means our concavity does not change

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