Answer to Question #346703 in Algebra for Dhruv bartwal

Question

1. Show that if 5/3< 2x< 11/3, then x∈{y∈R such that |y- 4/3| < 1/2}

2. Draw the graph of the function f, defined by f(x)= |x-6| + |5 - x| ; x∈ [2,8]

Answer 1

1.

Let's solve the inequalities.

$\frac{5}{3}< 2x< \frac{11}{3},$

we can divide by 2:

$\frac{5}{6}<x<\frac{11}{6},$

in interval notation: $x \in (\frac{5}{6}, \frac{11}{6}).$

$|y- \frac{4}{3}| < \frac{1}{2},$

$- \frac{1}{2}<y- \frac{4}{3} < \frac{1}{2},$

$- \frac{1}{2}+ \frac{4}{3}<y < \frac{1}{2}+ \frac{4}{3},$

$- \frac{3}{6}+ \frac{8}{6}<y < \frac{3}{6}+ \frac{8}{6},$

$\frac{5}{6}<y<\frac{11}{6},$

in interval notation: $y \in (\frac{5}{6}, \frac{11}{6}).$

We obtain the same intervals, so the first statement holds.

2.

$f(x)= |x-6| + |5 - x| ; x∈ [2,8]$

Let's consider this function on three intervals:

$f(x)=-(x-6) + (5 - x)=-x+6+5-x=$

$=-2x+11; x∈ [2,5)$

$f(x)=-(x-6) - (5 - x)=-x+6-5+x=1, x\in[5, 6)$

$f(x)=(x-6) - (5 - x)=x-6-5+x=$

$=2x-11, x\in[6, 8]$

Now we can draw the graph of the function f: