Roots of Absolute Valued Function (Hard)

Note That the function $f(x) = \dfrac{x}{4} + | | 10-x|-30|-10$ is non differentitable at $3$ points .

$f^{'}(x) = \dfrac{1}{4} - \dfrac{\left( |10-x|-30 \right) }{\left( | |10-x|-30| \right)} \cdot \dfrac{\left( 10-x \right)}{\left( |10-x| \right)}$

Our Critical points are $x = \hspace{0.5mm} -20 , 10 , 40$ .

corresponding $f(x)$ values are $-15,0,22.5$

So 3 pairs $\left(a,0\right)$ exist , one of them is $\left(10,0\right)$

Let's split it to cases:

$y=\frac{x}{4}+|20+x|-10\qquad if\quad x<10\\ y=\frac{x}{4}+|x-40|-10\qquad if\quad x>10$

Which can be further simplified:

$y=-\frac{3}{4}x-30\qquad if\quad x<-20\\ y=\frac{5}{4}+10\qquad if\quad -20<x<10\\ y=-\frac{3}{4}x+30\qquad if\quad 10<x<40\\ y=\frac{5}{4}-50\qquad if\quad 40<x\\$

Now we'll exemine the ends of our 4 linear intervals:

$y(-\infty)=\infty\\ y(-20)=-15\\ y(10)=22.5\\ y(40)=0\\ y(\infty)=\infty$

Which means we crossed 0 in the first interval, the second interval and on the point between the third and forth interval, all in all 3 points.

▄