Question

PLEASE HURRY:

Solve the equation:

|x+1|+|x-2|=3

(The brackets are absolute values)

Answer 1

`\underline{1.\, x\in(-\infty,-1]}\\-x-1+(-x+2)=3\\-x-1-x+2=3\\-2x=2\\x=-1\\\\\underline{2.\, x\in(-1,2]}\\x+1+(-x+2)=3\\x+1-x+2=3\\0=0\\\\x\in(-1,2\rangle\\\\\underline{3.\, x\in(2,\infty)}\\x+1+x-2=3\\2x=4\\x=2\\\\x\in\emptyset\\\\\\x=-1 \vee x\in(-1,2]\\\boxed{x\in[-1,2]}`

Answer 2

x ∈ [-1, 2]

Explanation:

The solution to the equation |x+1|+|x-2|=3 can be found by separately considering the domains where the function is defined differently.

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analysis

The absolute value function y=|x| has a vertex (turning point) at x=0. The translated function y=|x-h| will have its turning point at x=h. The sum of two absolute value functions with different turning points will have two turning points: one at each of the turning points of the constituents.

|x+1| has a turning point at x=-1

|x-2| has a turning point at x=2

This means there are three different domains to consider when determining solutions to the equation: (-∞, -1), [-1, 2), [2, ∞).

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(-∞, -1)

In this domain, the equation is equivalent to ...

-(x+1) -(x -2) = 3

-2x = 2 . . . . . . . . . simplify, add -1

x = -1 . . . . . . . . . . divide by -2, this x-value is not in the domain

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[-1, 2)

In this domain, the equation is equivalent to ...

(x +1) -(x -2) = 3

3 = 3 . . . . . true for all x in the domain

The solution is x ∈ [-1, 2).

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[2, ∞)

In this domain, the equation is equivalent to ...

(x +1) +(x -2) = 3

2x = 4 . . . . . . . . . . simplify, add 1

x = 2 . . . . . . . . . . . divide by 2. This x-value is in the domain

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consolidate solutions

Our solutions are ...

x ∈ [-1, 2) ⋃ [2, 2]

or ...

x ∈ [-1, 2]

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Additional comment

The graph shows the solution to the equation f(x) = 0, where the given equation is used to define the function f(x):

f(x) = |x +1| +|x -2| -3

f(x) will be zero for values of x that satisfy the original equation. We note that the solution is the closed interval [-1, 2], as we found above.