welding a piecewise function together at a point?

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?

h(x) = If(-4 < x < 2, 1 / 2 x + 5, x ≥ 2, x² - 7x + 16)

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

h(x) = If(-4 < x < 2, 1 / 2 x + 5, x ≥ 2, x² - 7x + 16)

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)