Question

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

Answer 1

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 change. 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!