Question

Find any relative extrema of the function. Use a graphing utility to confirm your result. \(f(x)=x \sinh (x-1)-\cosh (x-1)\)

Short answer

The function \(f(x)=x \sinh (x-1)-\cosh (x-1)\) has relative minima at \(x=0\) and \(x=2\), which have been confirmed using a graphical representation.

Step-by-step solution

Integral Calculation

The equation given is \(f(x)=x \sinh (x-1)-\cosh (x-1)\). First, calculate the first derivative \(f'(x)\) using the sum, product and chain rules. The derivative of \(x \sinh (x-1)\) is \(x \cosh (x-1) + \sinh (x-1)\) and the derivative of \(-\cosh (x-1)\) is \(\sinh (x-1)\). Thus, \(f'(x) = x \cosh (x-1) + \sinh (x-1) - \sinh (x-1) = x \cosh (x-1)\).

Find Critical Points

Set the derivative equal to zero and solve for \(x\) to find the critical points. So, \(x \cosh (x-1) = 0\). Solving for \(x\) gives \(x = 0, 2\). These are the critical points.

Classify the Critical Points

To classify these critical points, compute the second derivative \(f''(x)\) which is \(cosh(x-1) + x \sinh(x-1)\). Evaluate \(f''(x)\) at each critical point. If \(f''(x) > 0\), then \(x\) is a relative minimum. If \(f''(x) < 0\), then \(x\) is a relative maximum. \(f''(0) = cosh(0-1) = 1/cosh(1) > 0\), so \(x=0\) is a relative minimum. \(f''(2) = cosh(2-1) + 2sinh(2-1) = cosh(1) + 2sinh(1)\), and is greater than zero, so \(x=2\) is also a relative minimum.

Confirm Results with Graphing Tool

Now, confirm the results by graphing the function and checking to see that there are relative minima at \(x=0\) and \(x=2\). Use a graphing utility to do this.