Can someone help

[gloomy ember]

The i) question

[rose wedge]

Find the function f ?

[rose wedge]

Okay that was answered

[gloomy ember]

How do I Find the function?

@rose wedge

[rose wedge]

The condition

$f(\frac{x}{e}) \le lnx$

[weary axle]

kanny

[rose wedge]

Can you find any relation for f(x) from that?

[gloomy ember]

There is and 3rd condition

With domain x>0

It could be ln x, or e^x

Since there is e in conditions and an ln x

What do you think @rose wedge

[rose wedge]

How could it be e^x

[gloomy ember]

E^x

Has domain

x>0

Based on the inequality it asks to find f(x)

[rose wedge]

If it's e^x,
It should follow the inequality $f(x/e) \le ln(x)$

[weary axle]

kanny
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kanny

[rose wedge]

That's not true cuz at x = 0+, e^(x/e) > lnx

[gloomy ember]

So the only other option is ln x?

[weary axle]

kanny

[rose wedge]

I don't think so let's see?

[gloomy ember]

Yes

In ln x

It goes ln(x)-1

≤

ln x

And the same, ln x-1

@rose wedge

[rose wedge]

Huh?

Did you get f(x) ≤ ln(x) - 1?

[gloomy ember]

If you have f(x)=ln x

You get ln(x) -1≤ln x≤(ln x) -1

@rose wedge

[rose wedge]

Why did you substitute f(x) = ln x?

[gloomy ember]

To try if this is the function

[rose wedge]

Guess?

Ah okay

[gloomy ember]

Yes because of domain

And e

[rose wedge]

It's incorrect check the inequality on right

[gloomy ember]

I know

The which one should be the function

[rose wedge]

Did you get this point?

[gloomy ember]

No

[rose wedge]

Two functions f(x) and k(x) are related

[gloomy ember]

And how this will help?

[rose wedge]

By, f(x/a) = k(x)

It would help. . Try to understand it

[gloomy ember]

So f(x)=ex

[rose wedge]

Then f(x) = k(ax)

[gloomy ember]

?

The function is e*x?

[rose wedge]

No no

[gloomy ember]

Can you tell me fast cause I have to sleep asap

[rose wedge]

Man alright. I'd suggest you to try to understand these tho.

[gloomy ember]

Sure I do but not now

But in the inequality there isn't another function

To compare

And use that

[rose wedge]

$f(\frac{x}{e} \le ln(x)
f(x) \le ln(e.x)
f(x) \le ln(e) + ln(x)
f(x) \le 1 + ln(x)$

[gloomy ember]

Didn't understand

There are 2 inequality there

We have 3

[weary axle]

kanny

[rose wedge]

Sorry wait

[gloomy ember]

Ok

So f(x)=ln x+1?

[rose wedge]

f(x/e)≤ ln(x)
f(x) ≤ln(e.x)
f(x) ≤ ln(e) + ln(x)
f(x)≤ 1 + ln(x)

[rose wedge]

Yeah. .

If you solve the right inequality,
ln(x) + 1 ≤ f(x)

[gloomy ember]

Yes it could be

So I had to separate ln and e

That's all

[rose wedge]

f(x) is both lessorequal ln(x) + 1 $ and greaterorequal ln(x) + 1 $
Means f(x) = ln(x) + 1

[gloomy ember]

Both inequality has the same product

In order to be and equal

[rose wedge]

No, like f(x) is less or equal to a,
And f(x) is greater or equal to a.
Would mean f(x) = a.
Cuz ot can't be less and greater at the same time

[gloomy ember]

That's right

Thanks for your help, I have to sleep now, bye.