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Differential Calculus> in numerator first term is e raise to pow...

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in numerator first term is e raise to power integral part of x ….....please explain anyone please ...please...

milind , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$f(x) = \frac{e^{[x]} - e^{-x}}{e^{x}+e^{-x}}$
If it is e raise to power integral part of x, then
$f(0) = \frac{e^{[0]} - e^{-0}}{e^{0}+e^{-0}}$
$f(0) = \frac{1 - 1}{1+1}$
$f(0) = 0$
$f(-1) = \frac{e^{[-1]} - e^{-(-1)}}{e^{-1}+e^{-(-1)}}$
$f(-1) = \frac{e^{-1} - e^{1}}{e^{-1}+e^{1}}$
$f(-1) = \frac{e^{-1} - e}{e^{-1}+e}$
Its not equal to zero.
So there is something wrong with the question or solution you upload.
Please check it again.

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