Parker C. answered • 09/19/24
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I briefly explain each of the concepts you're being asked to understand in this video. I couldn't go into detail with the nuances and such due to hitting the max video length, but I'd be happy to answer any questions or such in this comment thread or in my direct messages! Hope this helps!
Parker C.
Glad to help! And no worries, there is A LOT to remember in calculus. So the first step is to find where the derivative is zero. In other words (as you put it) where f(x)=0. This gives you the locations where the derivative MIGHT change from positive to negative or vice versa. When the derivative changes from positive to negative, you have a local min, which doubles as a POSSIBLE absolute min. When the derivate changes from negative to positive, you have a local max, which doubles as a possible absolute max. Do not forget that both endpoints function as POSSIBLE absolute mins/maxes as well! So, write down all the possible mins and maxes you discover. Plug them back into the original A(x) formula and write the outcome. The final answer will be: The x-value(s) that give the lowest output along with the output (there may be multiple) and the x-value(s) that give the highest. Always make sure to check exactly what the question is asking! Some ask for the x value only. Some ask for the absolute min/max. This question seems to ask for both. If you need a refresher on how to find where the first derivative is negative/positive, message me and I can draw a little number line expressing it visually.
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09/19/24
Lily M.
Thank you so much! You explained it beautifully! Just one question... to find the max/min I just have to find the derivative and then find where f(x) = 0, right? It's been a little while since I've done this.09/19/24