 Try this beautiful problem on Graphs in Calculus, useful for ISI B.Stat Entrance.

## Graphs in Calculus | ISI B.Stat Entrance | Problem 698

Four graphs marked G1, G2, G3 and G4 are given in the figure which are graphs of the four functions $f_1(x) = |x – 1| – 1, f_2(x) = ||x –1| – 1|, f_3(x) = |x| – 1, f_4(x) = 1 – |x|$, not necessarily in the correct order.The correct order is

• (a) $G_2, G_1, G_3, G_4$
• (b) $G_3, G_4, G_1, G_2$
• (c) $G_2, G_3, G_1, G_4$
• (d) $G_4, G_3, G_1, G_2$

### Key Concepts

Calculus

Graph

Functions

But try the problem first…

Source

TOMATO, Problem 698

Challenges and Thrills in Pre College Mathematics

## Try with Hints

First hint

We take the each functions and express it in intercept form.we expand the mod i,e take the value once positive and once negetive .so we will get two equations and solve them,we will get the intersecting point also and draw the graph……..

Can you now finish the problem ……….

Final Step

$f_1(x) = |x – 1| – 1$

$(x-1)-1=y$

$\Rightarrow x-y=2$……………..(1)

$\Rightarrow \frac{x}{2} +\frac{y}{-2}=0$$\Rightarrow (2,0) ,(0,-2)$

And

$-(x-1)-1=y$

$\Rightarrow x+y=0$……..(2)

$\frac{x}{1}+\frac{y}{1}=0$$\Rightarrow (1,0),(0,1)$

Now if we draw the graph of (1) & (2) we will get the figure $G_2$ and the intersecting point is $(1,-1)$

Similarly we can draw the graphs for other functions………….

The second function is $f_2(x) = ||x –1| – 1|$ i.e $x-y=2$,$x=y$,$x+y=1$…which represents the two figure as given in $G_3$.

The third function $f_3(x) = |x| – 1$ which gives $x-y=1$ & $x+y=-1$..if we solve this two equations as first function then we will get $G_1$

The third function will gives the $G_4$ graph

Similarly we will draw the graph for all given functions….

Therefore ,the correct ans is (c)