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Differentiation q

In part b why not x>1? How did we reach x<1?

Original post by Presto
In part b why not x>1? How did we reach x<1?



Think about the graph of y=(x-1)(x-3), and where the curve is above the x-axis (which corresponds to y>0)
Reply 2
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BRAAAD
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Your working out seems perfectly fine to me, are you sure it's not correct?
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Presto
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Original post by Prasiortle
Think about the graph of y=(x-1)(x-3), and where the curve is above the x-axis (which corresponds to y>0)


Yeah Ik but is this the standard way? Shouldn't the equation lead to that too? Hope you get my qs
If you treat it like a normal inequality with a quadratic the left and the right of the graph is above the x-axis or where y=0.

The values of x must lie less than 1 or greater than 3 as the inequality suggests they move apart from one another. That is the correct way of doing it.
(edited 5 years ago)
Original post by Presto
Yeah Ik but is this the standard way? Shouldn't the equation lead to that too? Hope you get my qs


If you want to do it purely algebraically, notice that if the product of two numbers is positive, either both are positive or both are negative. So in this instance, the first case gives x-3>0 and x-1>0, so x>3 and x>1, but x>3 automatically implies x>1, so the combination just reduces to x>3. Similarly, the other case gives x-3<0 and x-1<0, so x<3 and x<1, so x<1. Putting the two cases together gives x>3 or x<1, as before.

While this method does work well for quadratics, it becomes a bit more cumbersome when dealing with higher-degree polynomials, and so in general I tend to recommend that people use a graphical approach.
Reply 6
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Presto
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Original post by Prasiortle
If you want to do it purely algebraically, notice that if the product of two numbers is positive, either both are positive or both are negative. So in this instance, the first case gives x-3>0 and x-1>0, so x>3 and x>1, but x>3 automatically implies x>1, so the combination just reduces to x>3. Similarly, the other case gives x-3<0 and x-1<0, so x<3 and x<1, so x<1. Putting the two cases together gives x>3 or x<1, as before.

While this method does work well for quadratics, it becomes a bit more cumbersome when dealing with higher-degree polynomials, and so in general I tend to recommend that people use a graphical approach.


Makes sense, thanks!

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