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# Dear Sir,For the function f(x)=X^x . Find the points at which it is1.Increasing.2. Decreasing .Hence Determine which of e^(pi)  , (pi)^ e  is greater .

SAGAR SINGH - IIT DELHI
879 Points
10 years ago

Dear student,

## Increasing Functions

A function is "increasing" if the y-value increases as the x-value increases, like this:

It is easy to see that y=f(x) tends to go up as it goes along.

What about that flat bit near the start? Is that OK?

• Yes, it is OK if you say the function is Increasing
• But it is not OK if you say the function is Strictly Increasing (no flatness allowed)

### Using Algebra

What if you can't plot the graph to see if it is increasing? In that case is is good to have a definition using algebra.

For a function y=f(x):

 when x1 < x2 then f(x1) ≤ f(x2) Increasing when x1 < x2 then f(x1) < f(x2) Strictly Increasing

That has to be true for any x1, x2, not just some nice ones you choose.

509 Points
10 years ago

f(x) = xx

d/dx [f(x)] = d/dx [ xx ]

f1(x)       =xx [ 1 + logx ]

f1(x) = 0 , when 1+logx = 0 ,  xx cannot be 0..

1+ logx = 0

logx = -1

x = 1/e

if x > 1/e , then function is increasing

if x < 1/e , then function is decreasing

increasing => (1/e , infinity)

decreasing=> (-infinity,1/e)

jagdish singh singh
173 Points
10 years ago

$(2)Here We have Use The Inequality \bf e^x>1+x\forall x>0(Prove Yourself)\\\\ Now Put x=\frac{\pi}{e}-1,We Get\\\\ \displaystyle e^{\frac{\pi}{e}-1}>\frac{\pi}{e}-1+1\Leftrightarrow \frac{e^{\frac{\pi}{e}}}{e}>\frac{\pi}{e}\\\\ So e^{\frac{\pi}{e}}>\pi\Leftrightarrow \boxed{\boxed{e^{\pi}>\pi^e}}$