# If #p(m) = 2m^2-4m+3#, what is #p(1/3)#?

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To find ( p\left(\frac{1}{3}\right) ), substitute ( m = \frac{1}{3} ) into the given function ( p(m) = 2m^2 - 4m + 3 ):

[ p\left(\frac{1}{3}\right) = 2\left(\frac{1}{3}\right)^2 - 4\left(\frac{1}{3}\right) + 3 ] [ p\left(\frac{1}{3}\right) = 2\left(\frac{1}{9}\right) - \frac{4}{3} + 3 ] [ p\left(\frac{1}{3}\right) = \frac{2}{9} - \frac{4}{3} + 3 ] [ p\left(\frac{1}{3}\right) = \frac{2}{9} - \frac{12}{9} + \frac{27}{9} ] [ p\left(\frac{1}{3}\right) = \frac{2 - 12 + 27}{9} ] [ p\left(\frac{1}{3}\right) = \frac{17}{9} ]

So, ( p\left(\frac{1}{3}\right) = \frac{17}{9} ).

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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.

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.

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.

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