Find the set of all values of x for which the inequality $\left | x-3 \right |+\left | x+2 \right |< 11$ holds

Question

I to go about this question, any help will be appreciated.

Answer 1

There are a couple of graphical ways to solve this.

Method 1. Consider $x,$ $3,$ and $-2$ as three points on the real number line, and construct a degenerate triangle with these three points as its three vertices. The perimeter of this triangle is $\lvert x-3\rvert + \lvert x+2\rvert + \lvert 3 - (-2)\rvert = \lvert x-3\rvert + \lvert x+2\rvert + 5.$ So the condition $\lvert x-3\rvert + \lvert x+2\rvert < 11$ is equivalent to the condition that the perimeter of the triangle must be less than $16.$

But the perimeter of a degenerate triangle is twice the distance between its extreme points. Imagine the perimeter as a rubber band around the points $-2,$ $3,$ and $x,$ and notice that the rubber band is stretched to its longest when $x$ is far less than $-2$ or far greater than $3.$ The perimeter would be exactly $16$ if $x$ is $8$ units to the left of $3$ (so $x=-5$) or $8$ units to the right of $-2$ (so $x = 6$). To have a perimeter less than $16,$ we can put $x$ anywhere between those two limits: $-5 < x < 6.$

Method 2. Plot the graphs of the two functions $y=\lvert x-3\rvert$ and $y = \lvert x+2\rvert$ and graphically take the sum of the two functions. We see that at $-2,$ the sum is $5,$ and to the left of $-2,$ the function plots are both lines with slope $-1,$ so the plot of the sum is a line of slope $-2$ through the point $(-2,5).$ Similarly, we find that to the right of $3$ the plot of the sum is a line of slope $2$ through $(3,5).$ Between $-2$ and $3$ the slopes cancel out, so we have a constant function between the points $(-2,5)$ and $(3,5).$ In short, we have a kind of U-shaped graph.

Now plot the function $y=11$ as a horizontal line. The plot of $\lvert x-3\rvert + \lvert x+2\rvert$ intersects this line somewhere to the left of $-2$ and somewhere to the right of $3,$ and is below the line everywhere in between. That is, the condition $\lvert x-3\rvert + \lvert x+2\rvert < 11$ means $x$ is between those two intersection points. Since the slope at the leftmost intersection is $-2,$ and the intersection is $6$ units above $(-2,5),$ it must be $3$ units to the left of $(-2,5),$ that is, at $(-5,11).$ Similarly we find that the rightmost intersection is $3$ units to the right of $(3,5),$ that is, at $(6,11).$ Therefore $-5 < x < 6.$

Answer 2

Let $f(x)=|x-3|+|x+2|$.

It's obvious that $f$ is a convex function.

Thus, the equation $f(x)=11$ has two roots maximum.

But $6$ and $-5$ are roots, which gives the answer $-5

Done!

Answer 3

you must do casework: for $$x\geq 3$$ we have $$x-3+x+2<11$$ for $$-2\le x<3$$ we have $$-(x-3)+x+2<11$$ and for $$x\le -2$$ we get $$-(x-3)-(x+2)<11$$ can you proceed?

Answer 4

If you wanted an algebraic way to do it, then since both sides of the inequality are positive:

$$\begin{align} |x-3|+|x+2|<11 \;&\Longrightarrow\; \left(|x-3|+|x+2|\right)^{2}<121 \\[0.1cm] &\Longrightarrow\; |x-3|^{2}+|x+2|^{2} + 2|x-3|\,|x+2| <121 \\[0.1cm] &\Longrightarrow\; (x-3)^{2}+(x+2)^{2} + 2|(x-3)(x+2)| <121 \\[0.1cm] &\Longrightarrow\;|x^{2}-x-6| < 54 -x^{2} + x \\ \end{align}$$

Then using $|x|

$$-54+x^{2}-x \,<\, x^{2}-x-6 \,<\, 54 -x^{2} + x$$

The first inequality: $-54+x^{2}-x \,<\, x^{2}-x-6$ is satisfied for all $x\in\mathbb{R}$ whereas the second $x^{2}-x-6 \,<\, 54 -x^{2} + x$ reduces to:

$$x^{2}-x-30 = (x+5)(x-6) \,<\, 0$$

which you should be able to solve. Admittedly, this is rather long as not as elegant, but it gets you there!