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# Justification using second derivative: maximum point

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.A (LO)
,
FUN‑4.A.10 (EK)
,
FUN‑4.A.11 (EK)

## Video transcript

we're told that given that H prime of negative four is equal to zero what is an appropriate calculus based justification for the fact that H has a relative maximum at X is equal to negative four so right over here we actually have the graph of our function H this is a graph y is equal to H of X and we don't have graphed the first derivative but we do have graphed the second derivative right here in this orange color H prime prime so they're telling us given that H prime of negative four is equal to zero so that's saying that given that the first derivative at x equals negative four is equal to zero and you can see that the slope of the tangent line when X is equal to negative four does indeed equal zero so given that what is a calculus based let me underline that a calculus based justification for the fact that H has a relative maximum at x equals negative four so this first one says that the second derivative at x equals negative four is negative now what does that tell us if the second derivative is negative that means that the first derivative is decreasing which is another way of saying that we are dealing with a situation where at least at x equals negative 4 we are concave downwards downwards which means that the general shape of our curve is going to look something like this around x equals negative 4 and if the slope at x equals negative 4 is 0 well that tells us that yes we indeed are dealing with a relative maximum point if the second derivative at Point was positive then we would be concave upwards and then if our derivative is 0 there would say ok that's a relative minimum point but this is indeed true the second derivative is negative at x equals negative 4 which means we are concave downwards which means that we are a upside-down U and that point where the derivative is 0 is indeed a relative maximum so let me sir that is the answer and we're done but let's just rule out the other ones H increases before x equals negative 4 that is indeed true before X equals negative four we are increasing and H decreases after it that is true and that is one rationale for thinking that hey we must have a maximum point assuming that our function is continuous at x equals negative four so this is true it is a justification for relative maximum but it is not calculus based and so that's why we can rule this one out the second derivative has a relative minimum at x equals negative 4 well it does indeed seem to be true there's a relative minimum there but that's not a justification for why this is why H of negative 4 or why we have a relative maximum at x equals negative 4 for example this you could have a relative minimum in your second derivative but your second derivative could still be positive there so what if the second derivative was like that that would still be a relative minimum but if it was positive at that point then you would be concave upwards which would mean that it x equals negative 4 or your original function wouldn't have a maximum point it would have a minimum point and so just a relative minimum isn't enough in order to know that you are dealing with a relative maximum you would have to know that the second derivative is negative there now this fourth choice H prime prime is concave up it does indeed look like the second derivative is concave up but that by itself does not justify that the original function is concave up for example well I could use this example right here this is a potential second derivative that is concave upwards but it is positive the entire time and if your second derivative is positive the entire time that means that your first derivative is increasing the entire time which means that your original function is going to be concave upwards to the entire time and so if you're concave upwards the entire time then you would not have a relative maximum at x equals negative 4 so we would rule that one out as well
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