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when they say f(x)=2 are they asking me to look on the x or y axis?
you have to intersect the graph of \(f(x)\) with the line \(y=2\), then you have to search for the x-coordinates of the intersection points
do you know if the desmos calculator can plug in piecewise functions?
I don't know, sorry :(
so what you saying is a horizontal line at y=2 has to go across and what ever it hits thats the number?
I see two different lines on the graph so what if it hits both of the graphs
you can solve your exercise, using the first graph above, here is how: |dw:1446572358214:dw|
therefore, you have to do the same procedure using the subsequent lines: \(y=4,\;y=5,\;y=8\)
each time you have to search for the values of \(x_1,\;x_2\)
how do you search for x1 an y1 using the graph above
when you get those two values then what do you do?
you have to use the millimeter paper
then what do you do with the two values doesn't the answer have to be one value
now, x=1.085 is less than 2, so x=1.085 is a solution, furthermore, x=3 is greater than 2, so x=3 is the other solution FInally, if we have no intersections, then our equation has no solutions If I use the line \(y=8\) you should get no solutions, please check
ok let me check
yeah it says 0
so I use the one on the left
oh so for f(x)=2 there is 2 solutions
correct! If \(f(x)=2\) we have 2 solutions, if \(f(x)=8\) we have no solutions
\(f(x)=8\) gives no solutions, since x=2.169 >2, and x=0.8 < 2
correct! we have 2 solutions
so f(x)=4 it is 2 solutions also
so does that mean f(x)=5 has two solutions also?
here: \(f(x)=5\) gives one solution, since x=1.5, is less than 2
the intersection with the blue lines, are acceptable, if the corresponding x-coordinate is greater or equal to 2, please refer to your definition of the piecewise function
I see so for these problems we also have to use the: 0 <= x < 2 2 <= x <= 4 along with the graph
so f(x)=2 has 2 solutions f(x)=4 has 2 solutions f(x)=5 has 1 solution f(x)=8 has 0 solutions I understand this better now I will study everything we talked about thank you for taking the time to explain this to me :)