How can we find out the interval in an inequality?I have the inequality: ( x + 2 ) ( x − 3 ) &lt; 0 . In thise case, I'm expected to find out two values by evaluation ( x + 2 ) = 0 and ( x − 3 ) = 0 and then divide the real number line in 4 intervals and figure out for which intervals the inequality holds by figuring out whether the expression on the LHS is less than 0 or greater than 0.

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How can we find out the interval in an inequality?I have the inequality:   (  x  +  2  )  (  x  −  3  )  &amp;lt;  0 . In thise case, I&#039;m expected to find out two values by evaluation   (  x  +  2  )  =  0 and   (  x  −  3  )  =  0 and then divide the real number line in 4 intervals and figure out for which intervals the inequality holds by figuring out whether the expression on the LHS is less than 0 or greater than 0.

Barbeyiae6a · Accepted Answer

Step 1For a continuous function f(x) on   [  a  ,  b  ] , let  f  (  a  )  &amp;gt;  0and  f  (  b  )  &amp;lt;  0or vice-versa. Then, by the intermediate value theorem, there exists   c  ∈  (  a  ,  b  ) such that   f  (  c  )  =  0 .So, when a continuous function changes sign, the graph must pass through the x-axis where its value is zero.Suppose   f  (  p  )  =  0 and   f  (  q  )  =  0 ,   p  &amp;lt;  q , such that   f  (  x  )  ≠  0 in (p, q). Then, in the interval (p, q), f must either be positive throughout or negative throughout. If this were not true, then, as seen above, there would be some   p  &amp;lt;  r  &amp;lt;  q such that   f  (  r  )  =  0 . So, the zeroes of f demarcate the interval into regions where it is either only positive or only negative. This makes it easy to check the sign by simply finding the sign of the function at any convenient point in the interval.

bulbareeh5kl · Accepted Answer

Step 1In high school we learn that     y  =  (  x  +  1  )  (  x  +  3  )  =      x    2    +  4  x  +  3   is the equation of a (vertical) parabola opening upwards, and thus it is negative (meaning     (  x  +  1  )  (  x  +  3  )  &amp;lt;  0   which is precisely when its graph is below the x− axis) exactly between its roots, meaning     y  &amp;lt;  0    ⟺    −  3  &amp;lt;  x  &amp;lt;  −  1   , and positive otherwise.Using this very beautiful relation between algebra and geometry makes things much simpler, in my opinion, and also more intuitive.For zero:     (  x  +  1  )  (  x  +  3  )  =  0   it is way simpler: we know that     a  b  =  0    ⟺    a  =  0   or     b  =  0    , that&#039;s why we do what you say we do to solve these equalities

How can we find out the interval in an inequality? I have the inequality: ( x + 2 )

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