Question

Let f(x)=| |e^x-1| -1| then f(x) is non differentlable for x belongs to:

Answer 1

Show your work, please. What have you done, so far?

Answer 2

@tkhunny I would like to know how to solve it algebraically also. Please, guide me. It doesn't matter if the Asker doesn't do his work. There are plenty of other students like me take advantage on other's stuff.

Answer 3

First, we must establish the meaning of the notation.

Is it \(e^{x-1}\;or\;e^{x}-1\)? The latter might be more interesting.

Answer 4

oh, i put |(e^x) -1| -1| on graphing tool to see the point the function is nondifferentiable but not know how to find it :)

Answer 5

Where is \(e^{x}-1 \ge 0\)?
Where is \(e^{x}-1 < 0\)?

Answer 6

when x > 0
when x <0

Answer 7

but |e^x -1| > 0 for all x but at x =0 we have a cups there.

Answer 8

y=|e^x-1|-1
try to plot its graph

Answer 9

You will get upward parabola from (0,-1)
And (ln2,0) in 1 st and 4 the quadrant
This part of graph is sufficient for the question
So, f(x) is non-differentiable for x belongs to {0,ln2} ,..that's what the question asking.

Answer 10

@Loser66

Answer 11

@tkhunny this is the work

Answer 12

https://www.desmos.com/calculator/r6uvvc16xs

Answer 13

only (0,1) and (0.693, 0) are two point the graph is nondiffentiable.

Answer 14

Yes and this is also the answer.

Answer 15

but I don't know how to solve it algebraically. We need it to back the logic up.
ok, let try one.
We need find the root of the function
||e^x -1 |-1| =0 when

Answer 16

Dude, solving these problems graphically is the best method.

Answer 17

I was not taught that!! to my courses, we have to prove it algebraically, graph is just to illustrate what we do.

Answer 18

But it saves your time OK lets do iit

Answer 19

How will you starstart

Answer 20

|e^x-1|-1=0

Answer 21

So |e^x-1 |= 1
Now two case
e^x-1=1. OR. e^x-1=-1
e^x=2. Or. e^x=0

Answer 22

I think you can solve from there

Answer 23

Do you need more help

Answer 24

You did not answer my questions.

Where is \(e^{x} − 1 \ge 0\) ?
Where is \(e^{x} − 1 < 0\) ?

No need to draw pictures. If you ALWAYS have to draw pictures, what will you do in 4 dimensions?

Answer 25

Every question have different ways to do....I do them in the way they suits me.

Answer 26

e^x-1>=0 when x>=0
And e^x-1<0 when x<0

Answer 27

@tkhunny

Answer 28

We have \(f(x) = ||e^{x}-1|-1|\)

When x >= 0, we have \(f_{1}(x) = |(e^{x}-1)-1| = |e^{x} - 2|\)

When x < 0, we have \(f_{2}(x) = |(1 - e^{x})-1| = |e^{x}| = e^{x}\)

We now KNOW something is funny at x = 0. This is a good place to check for differentiability.

Then we ask again,
when is \(e^{x} - 2 \ge 0\) and
when is \(e^{x} - 2 < 0\)?

We will find more suspicion associated with this answer.

Answer 29

x in reals only?

Answer 30

e^x-2>=0
e^x>=2
By taking ln we get x>=ln2

Answer 31

And e^x<2
When x<ln2 @tkhunny

Answer 32

Means something is also funny at x=ln2
So we should also check differentiability here
So we get two points x=0 and x=ln2

Answer 33

@tkhunny

Answer 34

Check it out! There is no reason to believe it is not differentiable elsewhere.

Feel free to write it out:

\(f(x) = \begin{cases}
e^{x} & x < 0 \\
2 - e^{x} & 0 \le x < \ln(2) \\
e^{x} - 2 & \ln(2) \le x \
\end{cases}
\)

Answer 35

This method is time consuming than graph isn't it?

Answer 36

I can't do the question fastly with this in exam

Answer 37

Can you taught me how to find no. of solution of a function?

Answer 38

That would depend on the speed of your thought. Think fundamentals FIRST - THEN worry about speed. Personally, in my college career, I never worried about speed. I concerned myself ONLY with fundamentals and I still managed to finish exams faster than most and with higher scores than most. If you are going to plan for speed, you must be very careful not to become sloppy.

Answer 39

You cannot find "ln(2)" on your graph. Will 0.693147 constitute a sufficient response? Maybe.

Answer 40

Sin^-1 x-|x-a|=0 ,is having at least one real root,then possible values of a can be?

Answer 41

But for entering a college, I have to crack an exam in which I got only 3 minutes for such questions.

Answer 42

Generally, one question per post. Always show YOUR work first - right up front.

Answer 43

OK wai

Answer 44

Easier to find if you are looking.
Also, some folks care about medals. You can give only one, no matter how long the thread.

Answer 45

I draw graph sin^-1 x..
Then y=a-x passed through point (1,π/2)

Answer 46

Can you have some good approache for it the way I tried is not much working

Answer 47

@tkhunny

Answer 48

What if I do like this
Sin ^-1(x)=|x-a|
-π/2<(x-a)>π/2 or -π/2<(a-x)>π/2

Answer 49

Inverse sine is limited to [-1,+1]
We're talking about staying withing '1' of 'a'.

Answer 50

What will I do next