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Help needed maths thx

  1. Mar 31, 2004 #1
    1how do i find the intersaction of sin(x) and [tex]\frac{2x}{\pi} [/tex]

    2 solve [tex]x-1 \leq \frac{1}{x-1}[/tex]
  2. jcsd
  3. Mar 31, 2004 #2


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    For the intersection, plot the two functions. That will give you one of the three answers right away. The other two are trickier.

    For the second, try first playing with [tex]y \leq \frac{1}{y}[/tex]. You can do it "by cases", or plotting the two sides of the equation to get an idea of the whole situation.
  4. Mar 31, 2004 #3
    oh i tried that but i got 4 answers that conflict each other, i dunno about the signs of them
  5. Mar 31, 2004 #4
    and for Q1 it wasnt able to solve algerbically only draw to graph in scale and measure it~
  6. Mar 31, 2004 #5


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    [tex]x-1 \leq \frac{1}{x-1}[/tex]

    If I remember correctly the algebraic way of doing this was to say:

    [tex]0 \leq \frac{1}{x-1} - x + 1[/tex]

    Make them the same fraction and find out where they change signs (where their linear factors in the numerator and denominator equal 0)
  7. Mar 31, 2004 #6
    Either way, you're going to have to multiply through by (x-1), which can be both negative or positive. The sign of (x-1) will determine which direction the inequality goes, so it is important.

    So for what values of x is (x-1) positive? Negative? What solutions do these two cases (+(x-1) and -(x-1)) yield?

    This is what ahrkron was getting at.

  8. Mar 31, 2004 #7
    [tex]x-1 \leq \frac{1}{x-1}[/tex]

    if x-1 greater than 0
    [tex](x-1)^2 \leq 1[/tex]
    [tex](x-1)^2-1 \leq 0[/tex]
    [tex]x^2-2x \leq 0[/tex]
    [tex]x(x-2) \leq 0[/tex]

    if x-1 less than 0

    [tex](x-1)^2 \geq 1[/tex]
    [tex](x-1)^2-1 \geq 0[/tex]
    [tex]x^2-2x \geq 0[/tex]
    [tex]x(x-2) \geq 0[/tex]

    conflict ~~~ =(
  9. Mar 31, 2004 #8
    i m now cureently study at sydney australia , the quailty there is so crap, my level of knowledge is dropping rapidly since came to sydney omg , the government do not really interested in high education, everything depends with us, we have to buy books past paper and do heaps more staff reqiued by school, and that only makes us standard compare to other country~ such a pity.
  10. Mar 31, 2004 #9
    ppl here must be older than me any suggest into my future career? thx

    i m losing my second place in maths to 6th, because the arrival of oversea studnets~

    what can i do to improve my self? many thx
  11. Mar 31, 2004 #10


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    The intersection of the graphs of y= sin(x) and [itex]y= \frac{2x}{\pi}[/itex] (it's the graphs that intersect, not the functions) is where [itex]sin(x)= \frac{2x}{\pi}[/itex]. There is no "algebraic" way to solve that but you could use any number of numerical methods: and, as ahrkron said, graphing it is simple. Plotting it on my handy-dandy TI-83 (and I remember when "four-function" calculators were big news!) I see immediately that (0,0) is one intersection and, since both functions are odd, there exist one point of intersection with x positive and one with x negative.
    Zooming in on the positive-x point, It looks like it is ([itex]\frac{\pi}{2}[/itex], 1).
    Well, yes, of course, sin([itex]\frac{\pi}{2}[/itex])= 1 and [itex]\frac{2}{/pi}(\frac{\pi}{2})= 1[/itex]. Graphing shows us that there exist exactly three points of intersection and simple calculation show that they are (0,0), ([itex]\frac{\pi}{2}[/itex], 1), and ([itex]-\frac{\pi}{2}[/itex], 1).

    As for the inequality: [tex]x-1 \leq \frac{1}{x-1}[/tex], I think the simplest general way to solve inequalities is to look at the equation:
    [itex]x-1 = \frac{1}{x-1}[/itex] is the same as (x-1)2= 1 so
    x-1= 1 => x= 2 or x-1= -1 => x= 0. Also the denominator of the fraction will be 0 when x-1= 0 => x= 1.
    The point of that is that inequality can "change" (from "<" to ">") only where the two sides are equal or where the function is discontinuous: in this case where the denominator is 0: here at 0, 1, and 2.

    If x= -1< 0, then the left side is (-1-1)= -2 and the right side is 1/(-2)= -1/2. -2< -1/2 so the inequality is true: therefore it is true for all x< 0.

    If x= 1/2 (between 0 and 1), the left side is (1/2-1)=-1/2 and the right side is -2. -1/2> -2 so the inequality is false for all x between 0 and 1.

    if x= 3/2 (between 1 and 2), the left side is (3/2-1)= 1/2 and the right side is 2. 1/2< 2 so the inequality is true for all x between 1 and 2.

    Finally, if x= 3 (larger than 2), the left side is (3-1)= 2 and the right side is 1/2. 3> 1/2 so the inequality is false for all x> 2.

    Since the problem was [itex]\leq[/itex], the equality is true at 0 and 2.

    The inequality is true for [itex]x\leq 0[/itex] and [itex]1< x \leq 2.
  12. Mar 31, 2004 #11


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    You don't have to multiply through by [itex]x-1[/itex] at all:

    [tex]x-1 \leq \frac{1}{x-1}[/tex]

    [tex]0 \leq \frac{1}{x-1} - (x - 1)[/tex]

    [tex]\frac{1}{x-1} - \frac{(x-1)^2}{x-1} \geq 0[/tex]

    [tex]\frac{1 - (x-1)^2}{x-1} \geq 0 [/tex]

    [tex]\frac{1- (x^2 - 2x + 1)}{x-1} \geq 0 [/tex]

    [tex]\frac{x^2 - 2x +1 -1}{x-1} \leq 0 [/tex]

    [tex]\frac{x(x-2)}{x-1} \leq 0[/tex]

    Critical values are [itex]x = 0,1,2[/itex] These are when they change signs, so as the final statement is true when [itex]x < 0[/itex] then:

    [tex]x \leq 0[/tex]


    [tex]1 < x \leq 2[/tex]

    See another way of doing it if you like :smile:
  13. Mar 31, 2004 #12
    omg how come all so smart~ thx all
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