How to Write (x+3)*|x-2| as a Piecewise Function?

  • Thread starter Thread starter Bostonpancake0
  • Start date Start date
  • Tags Tags
    Function
AI Thread Summary
To express (x+3)*|x-2| as a piecewise function, consider the behavior of the absolute value component, |x-2|, which changes at x=2. For x<2, |x-2| equals -(x-2), while for x≥2, |x-2| equals (x-2). Thus, the function can be defined as two separate expressions: for x<2, it simplifies to -(x+3)(x-2), and for x≥2, it simplifies to (x+3)(x-2). The absence of an absolute value around (x+3) does not affect the piecewise definition. Graphing both functions reveals distinct behaviors based on the value of x relative to 2.
Bostonpancake0
Messages
42
Reaction score
0
Write (x+3)*(absolute value of (x-2)) as a peice-wise defined function.

How do I set about doing this, considering (x+3) is not an absolute value??
 
Physics news on Phys.org
Try graphing both
y = (x + 3)|x - 2|
and
y = (x + 3)(x - 2) (without the absolute value)
by hand or by using a graphing utility. You will notice something when you compare the graphs.
 
ahhhh thank you
 
The fact that there is no absolute value around x+ 3 is irrelevant. There is an absolute value or x- 2 so look at x< 2, x> 2.
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Solve ##\left| x+3 \right|= \left| 2x+1\right|##
Replies
7
Views
3K
Why is ##h(x)=|x|## not a polynomial?
Replies
12
Views
1K
Is ##f(x)=2^{x}-1## considered an exponential function?
Replies
12
Views
2K
Confusion about ##\sqrt{x^2}= \left| x \right|##
Replies
1
Views
2K
Range of a function involving trigonometric functions
Replies
11
Views
2K
Find the range of the function ##f(x) = \frac{e^{2x}-e^x+1} {e^{2x}+e^x+1}##
Replies
8
Views
3K
Finding the domain (and range) of a logarithmic function
Replies
11
Views
3K
Query about how the domain of a binomial coefficient was calculated
Replies
2
Views
2K
To find the range of a given ##\sin## function
Replies
15
Views
2K
Need help in manipulating rational absolute value inequalities
Replies
11
Views
2K

Hot Threads

Recent Insights

Back
Top