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Math Help - Inequalities with variable in denominator

  1. #1
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    Inequalities with variable in denominator

    I have a question about inequalities like 1/x < 4. In this case the answer is (neg infinity, 0) U (1/4, pos infinity). It's intuitive that the set of negatives should be included, but the mechanical process of solving for x just goes out the window. That is, what rule do I break if i were to just solve for x... x>1/4 so the set would be simply (1/4, pos infinity). And, how do i know when the solving for x alone will not work.
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    Quote Originally Posted by eniuqvw View Post
    I have a question about inequalities like 1/x < 4. In this case the answer is (neg infinity, 0) U (1/4, pos infinity). It's intuitive that the set of negatives should be included, but the mechanical process of solving for x just goes out the window. That is, what rule do I break if i were to just solve for x... x>1/4 so the set would be simply (1/4, pos infinity). And, how do i know when the solving for x alone will not work.
    Hello,

    Ok, there is one thing you should know. When multiplying by a negative number, the direction of the inequality changes.

    For example : 2<3
    When multiplying by -1, you get -2>-3


    So here, intuitively, you should separate the situations :
    1) firstly, if x is positive
    2) secondly, if x is negative


    1) If x is positive :

    1/x<4

    Multiplying by x, we get : 1<4x, since x>0.
    --> x>1/4

    Solution to this is (1/4, pos infinity)

    2) If x is negative :

    Here, it's immediate because 1/x is negative, hence always inferior to 4.
    But let's apply the method.

    1/x<4

    Multiplying by x, we get 1>4x, because x is negative !

    --> x<1/4

    Because x is negative, you gotta find all the negative numbers which verify x<1/4.
    This is true for all x negative.

    Thus the solution to this is (neg infinity, 0)
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  3. #3
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    Quote Originally Posted by eniuqvw View Post
    I have a question about inequalities like 1/x < 4. In this case the answer is (neg infinity, 0) U (1/4, pos infinity). It's intuitive that the set of negatives should be included, but the mechanical process of solving for x just goes out the window. That is, what rule do I break if i were to just solve for x... x>1/4 so the set would be simply (1/4, pos infinity). And, how do i know when the solving for x alone will not work.
    Because my algebra is so poor I always solve inequalities by drawing graphs and then solving an equality.

    In your case I'd draw the graphs of y = 1/x and y = 1/4. Then I'd solve 1/x = 4. Then I'd use the graphs to get the solution to the inequality.
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    Quote Originally Posted by eniuqvw View Post

    I have a question about inequalities like 1/x < 4. In this case the answer is (neg infinity, 0) U (1/4, pos infinity).
    First Method: make 0 the RHS,

    \begin{aligned}<br />
   \frac{1}{x}&<4 \\ <br />
  \frac{1}{x}-4&<0 \\ <br />
  \frac{1-4x}{x}&<0.<br />
\end{aligned}

    To get a negative quantity we must study two cases:
    • If 1-4x<0 & x>0. Intersect these conditions and we have S_1=\left] \frac{1}{4},\infty  \right[.

    • If 1-4x>0 & x<0. Intersect these conditions and we have S_2=\bigg]-\infty,0  \bigg[.


    Final solution we'll be given by S=S_{1}\cup S_{2}=\bigg] -\infty ,0 \bigg[\cup \left] \frac{1}{4},\infty  \right[.


    Second Method: we obviously discard x=0, otherwise the initial inequality makes no sense. So we're gonna multiply both sides by x^2, since this quantity is positive, the inequality stays as it's written. Following up on this we have:

    \begin{aligned}<br />
   \frac{1}{x}&<4 \\ <br />
  x&<4x^{2} \\ <br />
  x(4x-1)&>0.<br />
\end{aligned}

    From here study two cases again:
    • If 4x-1>0 & x>0\implies S_{1}=\left] \frac{1}{4},\infty  \right[.

    • If 4x-1<0 & x<0\implies S_{2}=\bigg] -\infty ,0 \bigg[.


    Hence S=S_{1}\cup S_{2}=\bigg] -\infty ,0 \bigg[\cup \left] \frac{1}{4},\infty  \right[, which is we got before.
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  5. #5
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    A much easier approach is to multiply both sides by x^2 instead of x.

    As x^2 is always greater than 0 (x = 0 is not allowed in this case), there is no need to worry about the direction of the inequality sign.

    For 1/x < 4
    Multiplying both sides by x^2, we have
    x < 4x^2
    4x^2 - x > 0
    x(4x - 1) > 0

    The two roots of x(x/4 - 1) are x = 0 and x = 1/4.

    So the solution to the inequality is
    x < 0 or x > 1/4
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  6. #6
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    Then you're not gonna have trouble to check the second method above.
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