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welding a piecewise function together at a point?
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Hi I have these two functions f(x) and g(x).
f(x) = If(4 < x < 2, 1 / 2 x + 5)
g(x) =If(2 ≤ x, x²  7x + 16)
They graphically touch each other x = 2, but I can only get the program to acknowledge this if I change f(x) = If(4 <= x <= 2, 1 / 2 x + 5)
Any way (scriptwise) to force Geogebra to recognize that the functions do touch at the original condition for f eventhough f(x) = If(4 < x < 2, 1 / 2 x + 5)??
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Hi Mads!
The cut of the domains of f and g is empty.
In your first definition f has no value at x=2.
Greetings
mire2
Why not defining both as a single piecewise function?
Because I need to test if f(j) = g(j) or f'(j) not equel g'(j) when j = 2.
So is that possible?
Hi Mads! :smile:
Ok, another try. Yes, you can test it that way, but ...
f(j)=g(j) means, that j ∈ dom(f) ∩ dom(g) solving the equation
But in your given example dom(f) ∩ dom(g) = ∅, so there is no j solving the equation.
What you have to do is to proof if LimitBelow(f,2)==LimitAbove(g,2).
Edit: Hmm, that doesn't work, because the limits are "undefined", but with Simona's function h it works fine.
Greetings
mire2
try a single piecewise function and LimitAbove( <Function>, <Value> ) LimitBelow( <Function>, <Value> )
the correct syntax is
use it like in attached
Many thanks for your answer, Sir :)
Then my final questionaire is if I want to test whether (1 / 2 x + 5)' and (x²  7x + 16)' at x = 2 er different ? meaning
1 / 2 x + 5)' notequal (x²  7x + 16)' ?? and (1 / 2 x + 5)==(x²  7x + 16) ? At the given conditions.
Is that that possible?
type h'(x) or derivative(h) then use limitabove(h',2) and limitbelow(h',2)
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