If #f(x) = x^3 - 15/x#, what is #f(-1/3)#?

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Answer 1 · 2016-01-29T19:43:47

#f(-1/3) = 1214/27#

To compute #f(- 1/3)#, you need to plug #-1/3# for every occurence of #x# in your #f(x)# term:

#f(color(blue)(x)) = color(blue)(x)^3 - 15 / color(blue)(x)#

Replace each #color(blue)(x)# with #color(purple)(-1/3)#:

#f(color(purple)(-1/3)) = (color(purple)(-1/3))^3 - 15 / (color(purple)(-1/3))#

#= - 1^3 / 3^3 + 15 / (1/3)#

... to resolve the double fraction, remember that dividing by #1/3# is the same thing as multiplying with the reciprocal, namely #3/1 = 3#...

#= - 1 / 3^3 + 15 * 3#

#= - 1 / 27 + 45#

#= color(white)(x) 44 26/27 = 1214/27#,

whichever formulation you prefer.

Hope that this helped!