Is the Absolute Value Inequality $|4x-5|-|3x+1|+|5-x|+|1+x|=0.99 Solvable?

[anemone]

Show that the equation $|4x-5|-|3x+1|+|5-x|+|1+x|=0.99$ has no solutions.

 

[MarkFL]

My solution:

Spoiler

Let:

$$f(x)=|4x-5|-|3x+1|+|5-x|+|1+x|$$

We find that we may also write:

$$f(x)=\begin{cases}-3x+10, & x<-1 \\[3pt] -x+12, & -1\le x<-\dfrac{1}{3} \\[3pt] -7x+10, & -\dfrac{1}{3}\le x<\dfrac{5}{4} \\[3pt] x, & \dfrac{5}{4}\le x<5 \\[3pt] 3x-10, & 5\le x \\ \end{cases}$$

The graph of $f$ will have its minimum where the slope goes from negative to positive, thus we may conclude:

$$f_{\min}=f\left(\frac{5}{4}\right)=\frac{5}{4}$$

Hence:

$$f(x)=0.99$$

will have no real solution.

 

[anemone]

Good job, MarkFL! And thanks for participating!