Problems

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Solution:

1. Notice that \(f(x) = x - \tanh x\) has \(f'(x) = 1-\textrm{sech}^2 x = \tanh^2 x > 0\) so \(f(x)\) is strictly increasing on \((0, \infty)\) and \(f(0) = 0\) therefore \(f(x)\) is positive for all \(x\) positive
2. Let \(f(x) = x\sinh x-2\cosh x+2\) then \(f'(x) = \sinh x +x \cosh x - 2 \sinh x = x \cosh x -\sinh x = \cosh x ( x - \tanh x) > 0\) by the first part. \(f(0) = 0\) so \(f(x)\) is positive for all \(x\) positive.
3. Let \(f(x) = 2x\cosh2x-3\sinh2x+4x\) then \begin{align*} f'(x) &= 2\cosh 2x +4x\sinh 2x - 6 \cosh 2x + 4 \\ &= 4( x\sinh 2x-\cosh 2x +1) \\ &= 4(x2\cosh x \sinh x -2\cosh^2x ) \\ &= 8 \cosh^2 x (x - \tanh x) \end{align*} Which is always positive when \(x\) > 0, \(f(0) = 0\) so \(f(x) > 0\) for all positive \(x\).

Let \(f(x) = \frac{x(\cosh x)^{\frac{1}{3}}}{\sinh x}\) then \begin{align*} f'(x) &= \frac{(\cosh x)^{\frac13}\sinh x+\frac13 x \cosh^{-\frac23} x \sinh^2 x - x(\cosh x)^{\frac13} \cosh x}{\sinh^2 x} \\ &= \frac{\cosh x \sinh x + \frac13 x \sinh^2 x - x \cosh^2 x}{\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3\cosh x \sinh x + x( \sinh^2 x - 3 \cosh^2 x)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{\frac32 \sinh 2x + x( -2\cosh 2x - 2)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3 \sinh 2x -4x\cosh 2x - 4x}{6\cosh x^{\frac23} x \sinh^2 x} \\ \end{align*} which from the earlier part is always negative.