If #f(x)=7|x-2|#, find #f(0)#, #f(-2)#, #f(x+1)# and #f(x^2+2)#?

Answer 1

Genesis Allen

f(0) = 14 f(-2) = 28 f(x+1) = 7 |x-1| f(x^2+2) = 7 |x^2 - x + 4|

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Answer 2

Genesis Allen

#f(0)=14#, #f(-2)=28#, #f(x+1)=7|x-1|# and #f(x^2+2)=7x^2#

As #f(x)=7|x-2|#

#f(0)=7|0-2|=7xx|-2|=7xx2=14#

#f(-2)=7|-2-2|=7xx|-4|=7xx4=28#

#f(x+1)=7|x+1-2|=7|x-1|#

#f(x^2+2)=7|x^2+2-2|=7|x^2|=7x^2# *as #x^2# is always positive.

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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.

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