How do you solve and graph #abs(k-5)<=4#?

1 Answer
Apr 8, 2017

Answer:

#k in [1,9]#
enter image source here

Explanation:

Method 1: Split the absolute value possibilities

#{: ("possibility: "(k-5) < 0 (rarr k < 5),color(white)("X")"or"color(white)("X"),"possibility: "(k-5)>=0 (rarr k>=5)), ("then " abs(k-5)<=4,,"then " abs(k-5) <=4), (rarr 5-k <=4,,rarrk-5 <-4), (color(white)("XX")1 <= k,,color(white)("XX")k <=9), (k < 5" and "k>=1,color(white)("X")" or "color(white)("X"),k>=5" and "k<=9) :}#
Combining the two possibilities we have
#color(white)("XXX")1 <= k <=9#

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Method 2: square the two sides then check critical and sample interval points

Given #abs(k-5) <= 4#

Squaring both sides:
#color(white)("XXX")k^2-1ok+25 <= 16#

#color(white)("XXX")k-10k+9 <=0#

#color(white)("XXX")(k-1)(k-9) <= 0#

Giving the critical points #color(red)(k=1)# and #color(red)(k=9)#
We can then select arbitrary interval points:
for #k < 1#, #color(white)("XX")#I selected #color(blue)(k=0)#
for #k in (1,9)#, I selected #color(blue)(k=5)#
for #k > 9#, #color(white)("XX")#I selected #color(blue)(k=10)#

#{: (," | ",color(blue)(k=0),color(red)(k=1),color(blue)(k=5),color(red)(k=9),color(blue)(k=10)), (abs(k-5)," | ",5,4,0,4,5), ("? ? "<=4," | ",F,T,T,T,F) :}#

So the inequality is true for #k=1#, points between #k=1# and #k=9#, and for #k=9#

That is the inequality is valid for #k in [1,9]#

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Plotting is simply a matter of drawing this result on the number line.