Teorema di Lagrange 4
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
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Exercise:
Sfruttando il teorema del valor medio dimostra che una funzione derivabile f con f'geq è crescente.
Solution:
Per il Teorema del valor medio abbiamo che per ogni ab nell'insieme di definizione esiste c tale che fracfb-fab-af'cgeq e siccome b-ageq questo implica fbgeq fa ossia che la funzione è crescente. textbfIn alternativa: per assurdo se non fosse crescente avremmo due valori ab con fafb e in quel caso sempre per il Teorema del valore ermedio avremmo che esiste c con geq fracfb-fab-af'cgeq che è un controsenso
Sfruttando il teorema del valor medio dimostra che una funzione derivabile f con f'geq è crescente.
Solution:
Per il Teorema del valor medio abbiamo che per ogni ab nell'insieme di definizione esiste c tale che fracfb-fab-af'cgeq e siccome b-ageq questo implica fbgeq fa ossia che la funzione è crescente. textbfIn alternativa: per assurdo se non fosse crescente avremmo due valori ab con fafb e in quel caso sempre per il Teorema del valore ermedio avremmo che esiste c con geq fracfb-fab-af'cgeq che è un controsenso
Meta Information
Exercise:
Sfruttando il teorema del valor medio dimostra che una funzione derivabile f con f'geq è crescente.
Solution:
Per il Teorema del valor medio abbiamo che per ogni ab nell'insieme di definizione esiste c tale che fracfb-fab-af'cgeq e siccome b-ageq questo implica fbgeq fa ossia che la funzione è crescente. textbfIn alternativa: per assurdo se non fosse crescente avremmo due valori ab con fafb e in quel caso sempre per il Teorema del valore ermedio avremmo che esiste c con geq fracfb-fab-af'cgeq che è un controsenso
Sfruttando il teorema del valor medio dimostra che una funzione derivabile f con f'geq è crescente.
Solution:
Per il Teorema del valor medio abbiamo che per ogni ab nell'insieme di definizione esiste c tale che fracfb-fab-af'cgeq e siccome b-ageq questo implica fbgeq fa ossia che la funzione è crescente. textbfIn alternativa: per assurdo se non fosse crescente avremmo due valori ab con fafb e in quel caso sempre per il Teorema del valore ermedio avremmo che esiste c con geq fracfb-fab-af'cgeq che è un controsenso
Contained in these collections
| Title | Matched on |
|---|---|
| Teorema di Lagrange 8 | tagstitle |
| Teorema di Lagrange 9 | tagstitle |
| Teorema di Lagrange 10 | tagstitle |
| Teorema di Lagrange 11 | tagstitle |
| Teorema di Lagrange 12 | tagstitle |
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| Title | Matched on |
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| Teorema di Lagrange 10 | tagstitle |
| Teorema di Lagrange 11 | tagstitle |
| Teorema di Lagrange 12 | tagstitle |
| Teorema di Lagrange 1 | tagstitle |
| Teorema di Lagrange 2 | tagstitle |
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