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Stolb23 [73]
3 years ago
12

I need help on this question, I can't seem to understand piecewise functions!! ITS SO HARD.​

Mathematics
1 answer:
Sati [7]3 years ago
4 0

Answer:

f(x)\left \{ {{3x+1 if x\leq 0} \atop {-3x+1 if x>0}} \right.

Step-by-step explanation:

So if we first graph the given equation, we'll see the graph I've attached below.

Remember that piecewise functions are functions that change based on the circumstances. I know that sounds super confusing, but it's actually really simple!

In this case, for example, we see the line increasing from -∞ to 0, and then suddenly going downwards and decreasing. That's a good spot for us to notice because that indicates a <u>change</u>. We notice that the function looks different when x or x>0. If you break the function into those two parts, you see that they are just linear equations, but they're only visible when x is either greater than or less than 0.

Now that we notice this pattern, we can find the equation of the lines for both lines.

The points (-3,-8) and (-1,-2) are points on the first line, the one that increases (on the left). We can use those points to find the slope of the first line. Remember the slope equation:

m=\frac{y2-y1}{x2-x1}

Plug in your points:

m=\frac{-2-(-8)}{-1-(-3)}

m=\frac{6}{2}

m=3

So, the slope of the first line is 3. The y-intercept, looking at the graph, is 1. The equation of the first line is y=3x+1. We'll need this later.

Let's do the same thing for the second line. Just looking at the graph, we can see that this is the same exact line, just with a negative slope. So, the equation for the second line is y=-3x+1.

So now we can set up a piecewise function.

f(x)\left \{ {{3x+1} \atop {-3x+1}} \right.

The two functions in the bracket are the two different functions used in this graph. Now we need to figure out where each function is effective. Well, they share a y-intercept. Remember that a true function cannot have two points with the same x value. So the first function is effective to the left of x=0, while the second is effective to the right of x=0. In other words, when x\leq 0, f(x)=3x+1. But, when x>0, f(x)=-3x+1. Now our piecewise function looks like this:

f(x)\left \{ {{3x+1 if x\leq 0} \atop {-3x+1 if x>0}} \right.

And that is our piecewise function for the original function.

I know this is confusing, so please let me know if you have any questions! I hope this helps!

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