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+--- Thread: Graphical solving of systems of equations in the Function application. (/thread-18962.html)
Graphical solving of systems of equations in the Function application. - Jan 11 - 10-15-2022 06:31 AM
I still want to cover this topic in the Function app. This applies, of course, to finding graphical solutions to equations where the LN (x) and LOG (x) functions occur. I will give examples:
1. I solve graphically the system of equations: y = x ^ 2 -30 i y = LN (x)
We get 2 places of intersection. OKAY
2. Now he solves a graphically similar system of equations: y = x ^ 2 - 30 and y = LN (x - 2). Unfortunately, the calculator finds only one solution here, x = 5.59276. It does not show the second solution.
3. Now I change the function LN to LOG. Let's take a look at this.
y = x ^ 2 - 30 and y = LOG (x). We get 2 places of intersection. OKAY
4. Now we have: y = x ^ 2 -30 and y = LOG (x - 2). And here, unfortunately, as for LN, the calculator finds only one solution, x = 5.52697. It does not show the second solution.
Please correct this. And one more important note. All the above-mentioned systems of equations are solved well by the Advanced Graphing application.
RE: Graphical solving of systems of equations in the Function application. - Nigel (UK) - 10-15-2022 10:41 AM
The problem in these examples is that the root that the Graphing app doesn’t find is very close to 2: in your first example, the root is 2 + 5.109 x 10^(-12). In Home mode numbers on the Prime only have 12 digits, so this root cannot be found directly.
(I found the result above by replacing X by X+2 in each equation and solving for the new X.)
Nigel (UK)
RE: Graphical solving of systems of equations in the Function application. - Ioncubekh - 10-22-2022 08:02 AM
(10-15-2022 10:41 AM)Nigel (UK) Wrote: The problem in these examples is that the root that the Graphing app doesn’t find is very close to 2: in your first example, the root is 2 + 5.109 x 10^(-12). In Home mode numbers on the Prime only have 12 digits, so this root cannot be found directly.
(I found the result above by replacing X by X+2 in each equation and solving for the new X.)
Nigel (UK)
Nice explanation thanks