Domain of a function

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mantha1974

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Somehow this problem is probably a lot easier than I am making it out to be but I'm so stuck. I'm not sure where to begin.

Let f(x) = x^2 - |x-2|, find the domain.

My work:
x^2 - |x-2| = 0
x^2 = |x-2|
sqrt(x^2) = sqrt(|x-2|)

I'm not sure why I went to square roots, I think I'm just reaching for anything.
If anyone could help it would be great.. my brain is toast
 


Think about values that might be invalid for x. Are there any?

In this function, we don't need to be concerned about division by zero. We don't need to be concerned about taking the square root of a negative number.

If there are no restrictions on the value of x, then x can represent any Real number.

8-)

 
Awesome, thank you :)

If I were to solve f(x) = 0, I would set the function equal to zero or would I substitute 0 for the x's in the equation as in f(0)? (for the same function above)


Originally to solve f(x) = 0, this is the work I did:

f(0) = 0^2 - |0-2|
= -|0-2|
= -|-2|
= -2

Now thinking about it, I am unsure if I'm correctly following the directions.
 


It's important to understand function notation, when working with functions.

The symbolism f(x) is a variable.

In other words, f(x) is a number whose value varies when the value of x varies.

f(x) is just another way of writing the variable y, in an xy-coordinate system.

y = f(x)

f(x) = 0 means y = 0.

f(x) = 47 means y = 47.

f(x) = 350 means y = 350.

f(x) is a VARIABLE. Okay?

f(0) is NOT a variable. Once we replace x with an actual value (like zero), then the symbolism f(0) represents a CONSTANT.

f(0) is the constant we get after EVALUATING the function for x = 0.

f(-27) is the constant we get after evaluating the function for x = -27.

For example, y = 700 when x = -27.

Therefore, the constant f(-27) is 700.

So, we don't say "solve for f(0)". There is nothing to solve. We simply substitute 0 for x and do the arithmetic. We are not "solving"; we are "evaluating".

f(x) = 0

This is an equation that we can "solve".

Solving f(x) = 0 means to find the value(s) of x that cause y to equal zero.

The exercise you posted instructs neither for evaluating f(0) nor solving f(x) = 0.

Did you forget to type part of the exercise? :?

 
mantha1974 said:
… If I were to solve f(x) = 0 …
Click to expand...


OIC.

Substitute 0 for the variable f(x), and solve for x.

There are two solutions. 8-)

Hint: you can use the definition of absolute value to get rid of the absolute value symbols. In other words, rewrite the function for the two cases: x = 2, x < 2 and x > 2.

MY EDIT: Correct boundary error.
 
:(

I understand what you're saying. On any other problem I think I would be fine. What is throwing me is the x^2.

I tried: x^2 - (x-2) > 0 and x^2 - (x-2) < 0, but for some reason I can't get past this part... am I even rewriting the function for x < 0 and x > 0?

I really don't understand why I'm not getting this.
 
mantha1974 said:
Let f(x) = x^2 - |x-2|, find the domain.

my brain is toast
Click to expand...
Ola ,

\(\displaystyle f(x) = x^2 - | x - 2 |\)

Domain of f(x) : Any real number . . .

Brain Toasted with cheese and coffee yummyy ... :)
 
:) Thanks for that. My reply was the new problem. I don't think I was very clear in my explanation though so I'll type it out again real quick.

Let f(x) = x^2 - |x-2|

Solve f(x) =0

My best guess at this point is that my professor is asking for:

x^2 - |x - 2| = 0 ------- and to solve this equation.

My problem is that the x^2 is throwing me off and I can't get my brain to wrap around it. If it were a '2' or something, I would just move the abs to the other side (eliminating the - :D ) and set it < and > to the 2, solve for my two answers and be done... but i have a x^2.... :(
 


Hey there:

I just realized that I typed the wrong boundary point for taking cases. I apologize for the confusion this caused you. :oops:

We're trying to find the zeros of function f.

In other words, we're looking for values of x that cause f(x) to equal zero.

f(x) = x^2 - |x - 2|

To handle the absolute value expression, we consider the following three cases:

x < 2
x = 2
x > 2

Case x = 2 is easy.

f(2) does not equal zero, so x = 2 is not a zero of f.

Case x < 2:

If x is less than 2, then the expression x - 2 is negative.

In this case, by the definition of absolute value, |x - 2| = (-1)(x - 2) = -x + 2

Therefore, when x < 2, the polynomial defining function f becomes x^2 - (-x + 2).

f(x) = x^2 + x - 2, for x < 2

Case x > 2:

If x is greater than 2, then the expression x - 2 is positive.

In this case, by the definition of absolute value, |x - 2| = x - 2.

Therefore, when x > 2, the polynomial defining function f becomes x^2 - (x - 2).

f(x) = x^2 - x + 2, for x > 2

Solve each of these. From one, you'll get two Real solutions. For the other, you'll get two non-Real solutions.

The Real solutions are the zeros of function f; the non-Real solutions are not.

(When quadratic functions have Complex zeros containing the imaginary unit i, it indicates that the graph never reaches the x-axis.)

Let us know, if I wrote anything that you don't understand.
 
Ok, given the last explanation, this is what I did:


f(x) = x^2 - | x - 2 |, find the zeros of the function.

for x < 2

|x-2| = (-1)(x-2) = -x + 2
= x^2 - (-x +2)
= x^2 + x - 2

a = 1, b = 1, c = -2 (prep for quadratic formula)

[-(1) + sqrt( 1^2 - 4(1)(-2))] / 2(1)

= [ -1 + sqrt( 9 )] / 2

= ( -1 + 3 ) / 2

= ( -1 + 3 ) / 2 AND ( -1 - 3 ) / 2

( -1 + 3 ) / 2 = 2 / 2 = 1

AND

( -1 - 3 ) / 2 = -4 / 2 = -2

Therefore the zeros of f(x) are ( -2 , 0 ) AND ( 1 , 0 )

When I tried x > 2, the resulting answer was ( 1 + sqrt -7) / 2 <==== Non-real solution.


How does that look?
 
mantha1974 said:
… Therefore the zeros of f(x) are ( -2 , 0 ) AND ( 1 , 0 )

When I tried x > 2, the resulting answer was ( 1 + sqrt -7) / 2 <==== Non-real solution.

How does that look?
Click to expand...


Almost perfect! (Good job.) The numbers are correct, but the terminology needs adjusting.

The zeros of f are reported simply as -2 and 1.

We could also say that the solutions for f(x) = 0 are x = -2 or x = 1.

The ordered pairs (-2, 0) and (1, 0) are coordinates of the x-intercepts. 8-)

MY EDIT: Fixed typographical error.
 
mantha1974 said:
… When I tried x > 2, the resulting answer was ( 1 + sqrt[7] i) / 2 <==== Non-real solution.

How does that look?
Click to expand...


Good, except that we simplify the square root, and we need grouping symbols around the radicand, as shown. 8-)

 
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