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@terenzreignz @jim_thompson5910 @freckles Absolute minimum is lowest point on the graph?
at 5 the derivative goes from being negative (below the x axis) to positive (above the x axis) that means the function itself goes from decreasing to increasing
`Absolute minimum is lowest point on the graph?` correct @Ephemera
Ok so I would need to find the absolute minimum of f using the graph of f'?
you're in a calculus class, so naturally you should use calculus
find the local extrema and also test the endpoints
Wasn't taught that so you're gonna have to allow me a few minutes to google it.
hold on, I remember this problem a while back let me find it
Unless it's not needed.
haha you probably were, just that different wording was used :D Maximums or Minimums, they are collectively called Extremums Let's call a few points "extremum candidates" these are either endpoints of the interval of places where the DERIVATIVE is zero. I see four such candidates, can you identify them?
@jim_thompson5910 you helped me with a similar one a while back and I was still a little confused that's why I reposted.
take a look at my picture, it should be clear
derivative \(f'\) is negative, function\(f\) is decreasing (going dow) derivative is positive, function is increasing (going up) where it changes from deceasing to increasing you have a local minimum
(-2,-3) (2,0) (5,0) (6,4)
@satellite73 I see
Just the x-values will do.
Now, look at the values x = 2 and x = 5 Check how the derivative behaves around the said points. before x = 2, is the derivative positive or negative? what about after?
Ok got it. So I look at a point where the derivative is 0 and look at its surroundings to determine if it's a minimum or not.
So for a min before it would be negative and after positive while the max would be the opposite?
Yes. For it to be a minimum, it had better be decreasing BEFORE the x-value in question and increasing AFTER.
Yes, exactly :D
Gr8 8/8 m8 much appreci8
Not quite yet. We have only ever shown that a relative minimum exists on that x=5 point. Not yet an absolute minimum.
So... we can rule out x=2 as the function was decreasing both before and after x=2 What about the end points? The right end-point, x=6. Could the function possibly be lower at x=6 than it is at x=5?
No, As it is above the x-axis by a mile compared to x=5
"No" is correct but the justification is not quite there ^^ The reason is that the derivative has been positive for the entire duration on the interval (5,6]. That means the function has only gone up since then. What about the other endpoint, x=-1 ? Could the function be lower at x=-1 than it is at x=5?
Do you mean x=-2?
Sorry. Yes, I do mean x=-2 :)
Well it can't be because it was always negative at that point. ;)
Or the area surrounding that point is also negative.
haha the derivative has always been non-positive since it went from -2 to 5. that means the function has never increased since it was from -2 up until it got to 5. that means that, yes, the function is higher on -2 than it is on 5. So no, -2 is not the absolute minimum telling us conclusively that x=5 is at the ABSOLUTE MINIMUM of the function on the interval [-2 , 6] FINALLY DONE lol
8/8 m8 gr8 appreci8
No problem :D