Differential and integral calculus with a calculator?

Hmm. So in part, you can let the calculator reckon. (But really only Partial.

In particular at the zero points of the second derivation, it is already somewhat difficult to calculate the directly from calculator, since the calculator can only be derived numerically at individual points, and the calculator is also not designed for calculating the second derivation.

However, I would say that this task is easier and safer anyway without working the calculator.

In addition, as a rule: the solution path must be comprehensible. This may become more difficult if the calculator is excessively used.

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First of all, I will solve the task without a calculator for comparison. After that, I’ll go into how much the calculator can possibly take over…

Zero points of the first derivative:

Zero point of second derivation:

x=2 and x=4 are obviously simple zero points of the first derivative, so that there a sign change of the first derivative takes place. The first derivation describes an upwardly opened parabola which changes its sign from + to – at x=2 and changes its sign from − to + at x=4. Accordingly, x=2 is a (relative) maximum and x=4 is a (relative) minimum of f(x).

At x=3, f”'(x) obviously changes the sign from – to +, so that at x=3 there is a turning point of f(x).

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Now what the calculator can do…

The calculator can possibly be used to take over certain computing steps. For example, for the zero points of the first derivation, the square equation 3×2-18x+24=0 had to be solved, which could be left to the calculator calmly.

But maybe you want to take over more from the calculator. One can, for example, try to calculate the zero points of the first derivation directly without having to form the derivation itself beforehand.

For this, you can first enter the corresponding equation into the calculator…

The calculator is…

Then [SHIFT] [CALC] reaches the “SOLVE” command to solve the equation. However, the equation can be solved by the calculator only numerically, starting from a predetermined starting value. You have to tell the calculator what starting value you want to start. One can, for example, try to start with the start value x=0. So enter [0] and confirm [=].

Then that should look like you…

The calculator has thus found the solution x=2 of the equation.

However, this is only one of the two solutions that exist. For example, if you look into my solution path that I have described before, then I have come to x=4 as a second solution.

In order to find the second solution, it is necessary to work (instead of using the previous start value x=0) with another start value which is closer to the second solution x=4. You could now try different starting values. Next, I try for example with start value x = 5.

Result:

The calculator has now also found x=4 as a solution.

The problem that one has now, however, is there any other solutions that one has not found (due to the inappropriate choice of starting values)?

So you still have to try your brain yourself to find out: Are there any more solutions or have I found all the solutions?

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The next problem is: Is it about maximum or minimum positions?

There too, the calculator cannot help you directly.

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The next problem is: The zero points of the second derivative are also to be calculated. However, the calculator does not have the possibility to calculate the second derivation. [If one tries to put a first derivation into the first derivation, the calculator spits out a “syntax error”.]

Also there the calculator cannot help you directly.

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In the end, you should see that the calculator cannot simply solve the task directly.

But as an aid to take over certain partial calculations that occur during the solution path (for example, the resolution of the square equation 3×2 – 18x + 24 = 0),… Yes, for this you can use the calculator to solve the task.