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Dear Sir, For the function f(x)=X^x . Find the points at which it is 1.Increasing. 2. Decreasing . Hence Determine which of e^(pi) , (pi)^ e is greater .
Dear Sir,
For the function f(x)=X^x . Find the points at which it is
1.Increasing.
2. Decreasing .
Hence Determine which of e^(pi)  , (pi)^ e  is greater .


10 years ago

SAGAR SINGH - IIT DELHI
879 Points
							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.


10 years ago
509 Points
							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)

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

10 years ago
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• 51 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions