I need a bit of help........

G

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first one needing help on........... just dotnunderstand

Two statements, P and Q, about f(x) appear just below. Examine these statements and then choose the answer which best describes them.
f(x) = [g(x)]/[h(x)] where g(x) = x^4 - 2x^2 + 1 and h(x) = x^2 - 1 .

P: The domain of f(x) is all real x except x = 1 and x = -1.

Q: The graph of f(x) has no vertical asymptotes.

a. Only P is true.
b. Only Q is true.
c. Both P and Q are true.
d. Both P and Q are false


and second one.............

3x^4 + 5x^3 - 4x - 1 - (2x^4 - x^3 + x^2 - 7) = x4 + 4x3 - x2 - 4x - 8


 
you cannot divide by 0 so you need to look at the function in the denominator (h[x]) and solve for 0, that will give you the values at which x is undefined and thus, not in the domain.
 
Hello, achillesg22!

Careful . . . it's a trick question . . .

Two statements, P and Q, about f(x) appear just below.
Examine these statements and then choose the answer which best describes them.

\(\displaystyle f(x)\,=\,\frac{g(x)}{h(x)}\), where \(\displaystyle g(x)\,=\,x^4\,-\,2x^2 \,+\,1\) and \(\displaystyle h(x)\,=\,x^2\,-\,1\)

P: The domain of \(\displaystyle f(x)\) is all real \(\displaystyle x\) except \(\displaystyle x = 1\) and \(\displaystyle x = -1\)

Q: The graph of \(\displaystyle f(x)\) has no vertical asymptotes.

a. Only P is true . b. Only Q is true . c. Both P and Q are true . d. Both P and Q are false
Click to expand...
We have: \(\displaystyle \L\;f(x)\:=\:\frac{x^4\,-\,2x^2\,+\,1}{x^2\,-\,1}\:=\;\frac{(x^2\,-\,1)(x^2\,-\,1)}{x^2\,-\,1}\)

We know that \(\displaystyle x\) must not equal \(\displaystyle \pm 1\), but there are no vertical asymptotes!


If \(\displaystyle x\,\neq\,\pm1\), we can cancel and reduce the function to: \(\displaystyle \;f(x)\:=\:x^2\,-\,1\)

So we have a <u>parabola</u> with "holes" at \(\displaystyle x\,=\,\pm1\)
Code: 
        *           |           *
                    |
         *          |          *
      ----o---------+---------o--
         -1 *       |       * 1
               *    |    *
                    *
                    |
Therefore, both P and Q are true . . . answer (c).
 
soroban said:
So we have a parabola with "holes" at x = ±1.
Click to expand...
Note for the test: Using stuff like this, with those "holes" in 'em, is a common way to put a trick question on the test, the assumption being that you'll overlook this technical point.

I was in a class once where only one person noticed that on a test; it was very embarassing! :oops: :wink:

Hope that helps.

Eliz.
 
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