The graph of the function `y = g (x)` is shown.The number of solutions of the equation `||g(x)|-1|=1/2`, is

The graph of the function y=g(x) is shown.The number of solutions of the equation |g(x)|-1|=(1)/(2), is

The graph of the function y = g(x) is shown. The number of solutions of the equationm abs(absg(x)-1)=1/2 , is

The graph of the function y=f(x) is shown. Find the number of solutions to the equation ||f(x)|-1|=(1)/(2) .

The graph of the function y=f(x) is shown. Find the number of solutions to the equation ||f(x)|-1|=(1)/(2) .

The graph of the function y=f(x) is shown. Find the number of solutions to the equation ||f(x)|-1|=(1)/(2) .

The absolute valued function f is defined as f(x) = {{:(x,, x ge 0),(-x ,, x lt 0):}} and fractional part function g(x) as g(x) = x-[x], graphically the number of real solution(s) of the equation f(x) = g(x) is obtained by finding the point(s) of interaction of the graph of y = f(x) and y = g(x). The number of solutions (s) of |x-1| = {x}, x in [-1, 1] is

The absolute valued function f is defined as f(x) = {{:(x,, x ge 0),(-x ,, x lt 0):}} and fractional part function g(x) as g(x) = x-[x], graphically the number of real solution(s) of the equation f(x) = g(x) is obtained by finding the point(s) of interaction of the graph of y = f(x) and y = g(x). The number of solutions (s) of |x-1| = {x}, x in [-1, 1] is

The absolute valued function f is defined as f(x) = {{:(x,, x ge 0),(-x ,, x lt 0):}} and fractional part function g(x) as g(x) = x-[x], graphically the number of real solution(s) of the equation f(x) = g(x) is obtained by finding the point(s) of interaction of the graph of y = f(x) and y = g(x). The number of solution (s) |x-1| - |x+2| = k , when -3 lt k lt 3

The absolute valued function f is defined as f(x) = {{:(x,, x ge 0),(-x ,, x lt 0):}} and fractional part function g(x) as g(x) = x-[x], graphically the number of real solution(s) of the equation f(x) = g(x) is obtained by finding the point(s) of interaction of the graph of y = f(x) and y = g(x). The number of solution (s) |x-1| - |x+2| = k , when -3 lt k lt 3