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Question Number 21168 by Tinkutara last updated on 15/Sep/17

Let f(x) = ∣x − 1∣ + ∣x − 2∣ + ∣x − 3∣, then find the value of k for which f(x) = k has 1. no solution 2. only one solution 3. two solutions of same sign 4. two solutions of opposite sign

$$\mathrm{Let}\:{f}\left({x}\right)\:=\:\mid{x}\:−\:\mathrm{1}\mid\:+\:\mid{x}\:−\:\mathrm{2}\mid\:+\:\mid{x}\:−\:\mathrm{3}\mid, \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{k}\:\mathrm{for}\:\mathrm{which}\:{f}\left({x}\right) \\ $$$$=\:{k}\:\mathrm{has} \\ $$$$\mathrm{1}.\:\mathrm{no}\:\mathrm{solution} \\ $$$$\mathrm{2}.\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution} \\ $$$$\mathrm{3}.\:\mathrm{two}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{same}\:\mathrm{sign} \\ $$$$\mathrm{4}.\:\mathrm{two}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{opposite}\:\mathrm{sign} \\ $$

Answered by alex041103 last updated on 15/Sep/17

f(x)= { ((3x−6 , x∈(3, ∞), f(x)∈(3,∞))),((3, x=3)),((x , x∈(2, 3), f(x)∈(2,3))),((2, x=2)),((4−x , x∈(1, 2),f(x)∈(2,3))),((3, x=1)),((6−3x , x∈(−∞, 1), f(x)∈(3,∞))) :} ⇒f(x)∈[2,∞) ⇒1. k∈(−∞, 2) ⇒2. k=2 ⇒3. k=3 ⇒4. k∈[6, ∞)

$${f}\left({x}\right)=\begin{cases}{\mathrm{3}{x}−\mathrm{6}\:,\:{x}\in\left(\mathrm{3},\:\infty\right),\:{f}\left({x}\right)\in\left(\mathrm{3},\infty\right)}\\{\mathrm{3},\:{x}=\mathrm{3}}\\{{x}\:,\:{x}\in\left(\mathrm{2},\:\mathrm{3}\right),\:{f}\left({x}\right)\in\left(\mathrm{2},\mathrm{3}\right)}\\{\mathrm{2},\:{x}=\mathrm{2}}\\{\mathrm{4}−{x}\:,\:{x}\in\left(\mathrm{1},\:\mathrm{2}\right),{f}\left({x}\right)\in\left(\mathrm{2},\mathrm{3}\right)}\\{\mathrm{3},\:{x}=\mathrm{1}}\\{\mathrm{6}−\mathrm{3}{x}\:,\:{x}\in\left(−\infty,\:\mathrm{1}\right),\:{f}\left({x}\right)\in\left(\mathrm{3},\infty\right)}\end{cases} \\ $$$$\Rightarrow{f}\left({x}\right)\in\left[\mathrm{2},\infty\right) \\ $$$$ \\ $$$$\Rightarrow\mathrm{1}.\:{k}\in\left(−\infty,\:\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{2}.\:{k}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{3}.\:{k}=\mathrm{3} \\ $$$$\Rightarrow\mathrm{4}.\:{k}\in\left[\mathrm{6},\:\infty\right) \\ $$

Commented by Tinkutara last updated on 15/Sep/17

Answer of 3^(rd) part is wrong.

$$\mathrm{Answer}\:\mathrm{of}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{part}\:\mathrm{is}\:\mathrm{wrong}. \\ $$

Commented by alex041103 last updated on 15/Sep/17

f(x)= { ((3x−6 , x∈(3, ∞), f(x)∈(3,∞))),((3, x=3)),((x , x∈(2, 3), f(x)∈(2,3))),((2, x=2)),((4−x , x∈(1, 2),f(x)∈(2,3))),((3, x=1)),((6−3x , x∈(−∞, 1), f(x)∈(3,∞))) :} You can see that for x=1,3 f(x)=k=3 since sgn(1)=sgn(3) amd because the intervals in blue are open there aren′t any more x for which f(x)=3 Am I wrong?

$${f}\left({x}\right)=\begin{cases}{\mathrm{3}{x}−\mathrm{6}\:,\:{x}\in\left(\mathrm{3},\:\infty\right),\:{f}\left({x}\right)\in\left(\mathrm{3},\infty\right)}\\{\mathrm{3},\:{x}=\mathrm{3}}\\{{x}\:,\:{x}\in\left(\mathrm{2},\:\mathrm{3}\right),\:{f}\left({x}\right)\in\left(\mathrm{2},\mathrm{3}\right)}\\{\mathrm{2},\:{x}=\mathrm{2}}\\{\mathrm{4}−{x}\:,\:{x}\in\left(\mathrm{1},\:\mathrm{2}\right),{f}\left({x}\right)\in\left(\mathrm{2},\mathrm{3}\right)}\\{\mathrm{3},\:{x}=\mathrm{1}}\\{\mathrm{6}−\mathrm{3}{x}\:,\:{x}\in\left(−\infty,\:\mathrm{1}\right),\:{f}\left({x}\right)\in\left(\mathrm{3},\infty\right)}\end{cases} \\ $$$${You}\:{can}\:{see}\:{that}\:{for}\:{x}=\mathrm{1},\mathrm{3}\:{f}\left({x}\right)={k}=\mathrm{3} \\ $$$${since}\:{sgn}\left(\mathrm{1}\right)={sgn}\left(\mathrm{3}\right)\:{amd}\:{because} \\ $$$${the}\:{intervals}\:{in}\:{blue}\:{are}\:{open}\:{there}\:{aren}'{t} \\ $$$${any}\:{more}\:{x}\:{for}\:{which}\:{f}\left({x}\right)=\mathrm{3} \\ $$$${Am}\:{I}\:{wrong}? \\ $$

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