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    Hello I am doing C2- Differentiation and am finding values of x for which a function is increasing/ decreasing. When the factorised expression is shown to be smaller or equal to 0- when working it out, the inequality signs are not always right. It is not always possible to draw a graph with a hard function, so is there a way of understanding why the sign reverses?
    For example (x-2)(x-3)>0
    I'd usually expect to get x>2 , but it is actually x<2. Please may you possibly explain why?
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    (Original post by MathMeister)
    Hello I am doing C2- Differentiation and am finding values of x for which a function is increasing/ decreasing. When the factorised expression is shown to be smaller or equal to 0- when working it out, the inequality signs are not always right. It is not always possible to draw a graph with a hard function, so is there a way of understanding why the sign reverses?
    For example (x-2)(x-3)>0
    I'd usually expect to get x>2 , but it is actually x<2. Please may you possibly explain why?
    Of course it is always possible to sketch a graph. You are only dealing with quadratic inequalities.
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    (Original post by Mr M)
    Of course it is always possible to sketch a graph. You are only dealing with quadratic inequalities.
    Some are hard and have fractional indices and some I suppose turn into reciprocals.
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    So if the factorised equation is (x-3)(x-2)>0, pretend that the greater-than sign is just an equals sign soo... (X-3)(x-2)=0 then solve that giving x=3 and x=2.

    The way I learnt to do it is put it on a number line

    ___3_____2____
    Then put numbers into the gaps, so for example 5, 2.5, 1 and put those into the equation so for example 5, (5-3)(5-2) so 2x3 which is 6 > 0 so x could be >5 then 2.5 (2.5-3)(2.5-2) which it a negative number so won't work and finally 1 which would give a correct answer (>0) meaning that x<2 so the answer to this one would be x>3 or x<2 hoped this helped!

    Sorry if it didn't make complete sense but hopefully it helped
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    (Original post by MathMeister)
    Hello I am doing C2- Differentiation and am finding values of x for which a function is increasing/ decreasing. When the factorised expression is shown to be smaller or equal to 0- when working it out, the inequality signs are not always right. It is not always possible to draw a graph with a hard function, so is there a way of understanding why the sign reverses?
    For example (x-2)(x-3)>0
    I'd usually expect to get x>2 , but it is actually x<2. Please may you possibly explain why?
    (x-2)(x-3)>0 x=2 x=3 are where the graph crosses the x axis the function needs to be bigger than 0 and when the graph crosses the x axis it is below 0 so anything in the range 2-3 doesn't satisfy the inequality so x<2 and x>3 are the set of solutions.
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    (Original post by MathMeister)
    Hello I am doing C2- Differentiation and am finding values of x for which a function is increasing/ decreasing. When the factorised expression is shown to be smaller or equal to 0- when working it out, the inequality signs are not always right. It is not always possible to draw a graph with a hard function, so is there a way of understanding why the sign reverses?
    For example (x-2)(x-3)>0
    I'd usually expect to get x>2 , but it is actually x<2. Please may you possibly explain why?
    Why do you "expect to get x>2"? There are 2 ways of multiplying 2 quantities together to get a positive result:

    both brackets are positive
    OR
    both brackets are negative

    You need to deal with both cases that arise.
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    (Original post by MathMeister)
    ...
    For these sorts of questions, consider critical points. This is analogous to looking at the roots. I.e., suppose you had the quadratic

    x^2 + 5x + 6 = 0

    By factorising, and inspecting the roots, you can see that they are -2, -3. Now, consider the graph of this quadratic. If you're looking for the set of points such that the function is increasing, you'll notice the function increases to the right of the larger root and to the left of the smaller root. Therefore, for this question in particular, the solutions would be

    x &gt; -2 and x &lt; -3.

    Try and apply this reasoning to your questions. It is always helpful to envisage the graph.
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    (Original post by Indolence)
    For these sorts of questions, consider critical points. This is analogous to looking at the roots. I.e., suppose you had the quadratic

    x^2 + 5x + 6 = 0

    By factorising, and inspecting the roots, you can see that they are -2, -3. Now, consider the graph of this quadratic. If you're looking for the set of points such that the function is increasing, you'll notice the function increases to the right of the larger root and to the left of the smaller root. Therefore, for this question in particular, the solutions would be

    x &gt; -2 and x &lt; -3.

    Try and apply this reasoning to your questions. It is always helpful to envisage the graph.
    Hmm, a guy with a naruto sig/avy, good at maths and uses latex... Where have I seen that before?
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    (Original post by Dilzo999)
    ...
    Heh-heh. I'm a new user here, but I've friends who've showed me around, hence why I'm familiar with TSR. I've used MikTex (one of many LaTeX distributions) before.
 
 
 
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