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In order to deal with the modulus sign, we must take account of 2 possible cases:

Case 1: |2x + 1| = (2x +1). In this case we can solve algebraicly, preserving the inequality sign, to get...

To solve this problem, we must first differentiate:

Identify that we are able to use the product rule as our expression is of the form y = f(x)g(x) where f(x) = e^(2- x) and g(x) = ln(3x-...

most people are immediately confused when they see a surd or a fraction because they find it hard to apply the rule of integration directly to them. however it is not as complicated as it looks. all th...

Well, you need to have a clear mind and know well the pattern that they will gonna ask. Then all you have to do is make sure you understand all the skills that you required to need.

Whilst it is possible to do it algebraically, it's usually easier to solve it graphically. For example: for which values of x is |x+2| = |3x-6|. By sketching the graphs of y=|x+2| and y=|3x-6|, it is e...