Question

\(f(x)=\frac{x}{1+e^{1 / x}}\) \(f(x)\left(1+e^{1 / x}\right)-x=0\) \(f^{\prime}(x)\left(1+e^{1 / x}\right)-\frac{f(x) e^{l / x}}{x^{2}}-1=0\) \(x^{2} f^{\prime}(x) \frac{(x)}{f(x)}-f(x) e^{1 / x}-x^{2}=0\)

Short answer

Question: Verify that for the given function \(f(x) = \frac{x}{1+e^{1 / x}}\), the following equalities hold: \(f(x)(1+e^{1/x})-x=0\) and \(x^2 f'(x) \frac{x}{f(x)} - f(x)e^{1/x} - x^2=0\). Answer: We first verified that the given equation \(f(x)(1+e^{1/x})-x=0\) holds true by substituting the function \(f(x)\) into the equation and simplifying. Then, we differentiated the function \(f(x)\) with respect to x and found \(f'(x)\). Finally, we substituted the expressions for \(f(x)\) and \(f'(x)\) into the second equation and simplified to confirm that \(x^2 f'(x) \frac{x}{f(x)} - f(x)e^{1/x} - x^2=0\) also holds true. Both given equalities are verified.

Step-by-step solution

Verify given equation

First, we need to check if \(f(x)(1+e^{1/x})-x=0\). Let's substitute the function \(f(x)\) into the equation and simplify. \(f(x)(1 + e^{1/x}) - x = \frac{x}{1+e^{1 / x}} (1 + e^{1/x}) - x\) Now, we distribute the terms and simplify: \(\frac{x(1 + e^{1/x})^2}{(1 + e^{1/x})} - x = \frac{x(1 + e^{1/x}) - x(1 + e^{1/x})}{(1 + e^{1/x})} = 0\) Thus, the given equation is true.

Differentiate the function

Now, we need to differentiate the function \(f(x)\) with respect to x: \(f'(x) = \frac{d}{dx}\left(\frac{x}{1 + e^{1/x}}\right)\) Using the quotient rule, we get \(f'(x) = \frac{(1 + e^{1/x})\frac{d}{dx}(x) - x\frac{d}{dx}(1 + e^{1/x})}{(1 + e^{1/x})^2}\) Simplify and compute the derivatives: \(f'(x) = \frac{(1 + e^{1/x})(1) - x\left(-\frac{1}{x^2}e^{1/x}\right)}{(1 + e^{1/x})^2}\) Now we must simplify f'(x): \(f'(x) = \frac{1+e^{1/x}+\frac{x e^{1/x}}{x^2}}{(1+e^{1/x})^2}\)

Verify the second equation

The second equation to verify is: \(x^2 f'(x) \frac{x}{f(x)} - f(x)e^{1/x} - x^2=0\). Let's plug in \(f(x)\) and \(f'(x)\) and simplify: \(x^2 \left(\frac{1+e^{1/x}+\frac{x e^{1/x}}{x^2}}{(1+e^{1/x})^2}\right) \frac{x}{\frac{x}{1+e^{1/x}}} - \frac{x}{1+e^{1/x}}e^{1/x} - x^2\) Simplify by canceling terms: \(\left(\frac{x^3(1+e^{1/x}+\frac{x e^{1/x}}{x^2})}{(1+e^{1/x})}\right) - \frac{x}{1+e^{1/x}}e^{1/x} - x^2\) Now distribute the \(x^3\) term: \(\left(\frac{x^3+ x^2 e^{1/x} + x e^{1/x}}{(1+e^{1/x})}\right) - \frac{x}{1+e^{1/x}}e^{1/x} - x^2\) Observe that the first term in the numerator cancels with the last term \(-x^2\): \(\frac{x^2 e^{1/x}+ xe^{1/x}}{1+e^{1/x}} - \frac{xe^{1/x}}{1+ e^{1/x}} =0\) Factor out \(xe^{1/x}\): \(\frac{xe^{1/x}(x+1)}{1+e^{1/x}} - \frac{xe^{1/x}}{1+ e^{1/x}}\) Finally, observe that the numerator now exactly cancels, and we are left with 0: \(0 = 0\) Thus, the given second equation is also true.