Turning points

burgerandcheese

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Q: Find the coordinates of each of the turning points of the curve y = x +(8 - x²) and determine whether it is a maximum or minimum point.

I did dy/dx = 0 and I got x = ±2 , but x = -2 is extraneous so the curve only has a turning point at x = 2.
I graphed it on too:
https://www.desmos.com/calculator/lgyw282qoq

However, when I checked on wolframalpha it said the curve has a global maximum at x = 2 and a global minimum at x = -2√2
https://www.wolframalpha.com/input/?i=y%3Dx%2Bsqrt(8-x%5E2)

how do I get x = -2√2 ?
 
Global extrema don't necessarily occur only at turning points; they can occur at endpoints in the function's domain, also.

Look at the Desmos graph. The given function has only one turning point, so you don't need to locate any other points for this exercise.

You reported the x-coordinate of the turning point. Calculate the y-coordinate, too, and then report the turning point using (x,y) coordinates. Don't forget to state that the turning point is a local maximum.

Regarding global extrema, the turning point's y-coordinate is the global maximum. Therefore, the local maximum is also the global maximum. However, the function's global minimum is the y-coordinate of the lowest point on the graph, and we can see that it occurs at the smallest value of x in the domain. But again, the exercise doesn't ask about global extrema. (The largest value of x in the domain also leads to a local minimum, but it's not the global minimum.)

By the way, do you know how to determine whether a turning point occurs at a local max or min, without looking at a graph? :cool:
 
Global extrema don't necessarily occur only at turning points; they can occur at endpoints in the function's domain, also.

Look at the Desmos graph. The given function has only one turning point, so you don't need to locate any other points for this exercise.

You reported the x-coordinate of the turning point. Calculate the y-coordinate, too, and then report the turning point using (x,y) coordinates. Don't forget to state that the turning point is a local maximum.

Regarding global extrema, the turning point's y-coordinate is the global maximum. Therefore, the local maximum is also the global maximum. However, the function's global minimum is the y-coordinate of the lowest point on the graph, and we can see that it occurs at the smallest value of x in the domain. But again, the exercise doesn't ask about global extrema. (The largest value of x in the domain also leads to a local minimum, but it's not the global minimum.)

OK. Noted!

By the way, do you know how to determine whether a turning point occurs at a local max or min, without looking at a graph? :cool:

Yes, by using second derivative. Apart from that, I'm guessing we do completing the square and see how the numbers would behave...(I don't know how to explain it). Correct? Are there any other methods?
 
… by using second derivative … any other methods?
The second-derivative test always works.

In most cases, one could also evaluate the function on each side of the turning point, to check whether these values are both higher or both lower than the output at the turning point. For example, evaluate the function at three points (x=1.9999, x=2 and x=2.0001). If the output at x=2 is larger than the outputs on either side, then the point in the middle must be a maximum.

I would guess that your instructor expects to see the second-derivative test used. The other method might be considered a 'quick-and-dirty' approach, and it won't always work. :cool:
 
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For example, evaluate the function at three points (x=1.9999, x=2 and x=2.0001). If the output at x=2 is larger than the outputs on either side, then the point in the middle must be a maximum.
Although it will be true for almost all textbook problems it is still a dangerous approach. After all you can have crazy things happening between x=1.9999 and x=2.0001 like 1800 max and mins and many points of inflections.
 
The second-derivative test always works.

In most cases, one could also evaluate the function on each side of the turning point, to check whether these values are both higher or both lower than the output at the turning point. For example, evaluate the function at three points (x=1.9999, x=2 and x=2.0001). If the output at x=2 is larger than the outputs on either side, then the point in the middle must be a maximum.

Yes, that was what I learnt first to find maximum/minimum points, before being introduced to second derivatives :D
 
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