anonymous
  • anonymous
The figure below shows the graph of f ′, the derivative of the function f, on the closed interval from x = −2 to x = 6. The graph of the derivative has horizontal tangent lines at x = 2 and x = 4. Find the x-value where f attains its absolute minimum value on the closed interval from x = −2 to x = 6. Justify your answer.
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
https://gyazo.com/2c47b68b177653bc9680c8043feceba6
anonymous
  • anonymous
@terenzreignz @jim_thompson5910 @freckles Absolute minimum is lowest point on the graph?
anonymous
  • anonymous
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

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anonymous
  • anonymous
|dw:1449710951722:dw|
jim_thompson5910
  • jim_thompson5910
`Absolute minimum is lowest point on the graph?` correct @Ephemera
anonymous
  • anonymous
Ok so I would need to find the absolute minimum of f using the graph of f'?
jim_thompson5910
  • jim_thompson5910
you're in a calculus class, so naturally you should use calculus
jim_thompson5910
  • jim_thompson5910
find the local extrema and also test the endpoints
anonymous
  • anonymous
Say what?
anonymous
  • anonymous
Wasn't taught that so you're gonna have to allow me a few minutes to google it.
jim_thompson5910
  • jim_thompson5910
hold on, I remember this problem a while back let me find it
anonymous
  • anonymous
Unless it's not needed.
terenzreignz
  • terenzreignz
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?
anonymous
  • anonymous
@jim_thompson5910 you helped me with a similar one a while back and I was still a little confused that's why I reposted.
anonymous
  • anonymous
take a look at my picture, it should be clear
anonymous
  • anonymous
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
anonymous
  • anonymous
(-2,-3) (2,0) (5,0) (6,4)
jim_thompson5910
  • jim_thompson5910
anonymous
  • anonymous
@satellite73 I see
terenzreignz
  • terenzreignz
Just the x-values will do.
terenzreignz
  • terenzreignz
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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
So for a min before it would be negative and after positive while the max would be the opposite?
terenzreignz
  • terenzreignz
Yes. For it to be a minimum, it had better be decreasing BEFORE the x-value in question and increasing AFTER.
terenzreignz
  • terenzreignz
Yes, exactly :D
anonymous
  • anonymous
Gr8 8/8 m8 much appreci8
terenzreignz
  • terenzreignz
Not quite yet. We have only ever shown that a relative minimum exists on that x=5 point. Not yet an absolute minimum.
terenzreignz
  • terenzreignz
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?
anonymous
  • anonymous
No, As it is above the x-axis by a mile compared to x=5
terenzreignz
  • terenzreignz
"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?
anonymous
  • anonymous
Do you mean x=-2?
terenzreignz
  • terenzreignz
Sorry. Yes, I do mean x=-2 :)
anonymous
  • anonymous
Well it can't be because it was always negative at that point. ;)
anonymous
  • anonymous
Or the area surrounding that point is also negative.
terenzreignz
  • terenzreignz
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
anonymous
  • anonymous
Now 8/8?
anonymous
  • anonymous
8/8 m8 gr8 appreci8
terenzreignz
  • terenzreignz
No problem :D

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