Solution assignment 10 Inequalities with fractions

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Assignment 10

Solve:

$\displaystyle\frac{e^{x}-1}{e^{x}}>\displaystyle\frac{1}{e^{x}+1}$

Solution

Both denominators cannot be equal to $0$ for any value of $x$ . So, the inequality is valid for all $x$ .
We take the right-hand side to the left:

$\displaystyle\frac{e^{x}-1}{e^{x}}-\displaystyle\frac{1}{e^{x}+1}>0$

We convert fractions to the same denominator and get:

$\displaystyle\frac{e^{x}-1}{e^{x}}.\displaystyle\frac{e^{x}+1}{e^{x}+1}-\displaystyle\frac{1}{e^{x}+1}.\displaystyle\frac{e^{x}}{e^{x}}>0$

$\displaystyle\frac{e^{2x}-e^{x}-1}{e^{x}(e^{x}+1)}>0$

The factors in the denominator are positive for all $x$ (exponential functions) and thus the inequality is valid if the numerator is greater than $0$ . Thus:

$e^{2x}-e^{x}-1>0$

The numerator in the left-hand side is a quadratic function in $e^{x}$ and the solutions are ( $abc$ -formula):

$e^{x_{1,2}}=\displaystyle\frac{1\pm\sqrt{5}}{2}$

The graph of the quadratic function is an 'opens up' parabola and is positive if:

$e^{x}<\displaystyle\frac{1-\sqrt{5}}{2}$

or:

$e^{x}>\displaystyle\frac{1+\sqrt{5}}{2}$

The first inequality is not possible because the right-hand side is negative and the left-hand side is an exponential function which is always positive.
Thus we use the logarithm in the left- and right-hand side and get:

$x>\ln(\displaystyle\frac{1+\sqrt{5}}{2})$

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Solution assignment 10 Inequalities with fractions

Assignment 10

Solution

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