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How do I find all the critical points of #f(x)=x^2+4x-2#?

Answer 1

You find the critical points of a function by setting its first derivative to zero. In this case, you only need to remember the rule for deriving a power, which is #d/{dx} x^n = nx^{n-1}#. Also, you have to remember that the derivative of a sum is the sum of the derivatives, which means that we can derive this polynomial term by term.

So, deriving #x^2# gives #2x#, deriving #4x# gives 4, and deriving #-2# (as any other constant) gives zero. The first derivative of #x^2+4x-2# is thus #2x+4#.

Of course, #2x+4=0 \iff 2x=-4 \iff x=-2#.

The (only) critical point of your parabola is #(-2, -6)#, which is its vertex.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.