Find the local maximum and minimum values of the function $f(x)=-x-\frac{25}{x}$ using the Second Derivative Test.
Enter the points in increasing order of their $x$-coordinates.
Enter the first point in the form $(x, y)$ :
The sign of $f^{\prime \prime}(x)$ at this point is Click for list
Based on your answer, the conclusion is Cick for List
Enter the second point in the form $(x, y)$ :
The sign of $f^{\prime \prime}(x)$ at this point is cick for List
Based on your answer, the conclusion is Chick for Last
Answer
Check the answer: The function \(f(x)\) is continuous and differentiable for all \(x\neq 0\). The Second Derivative Test is valid for \(x=\pm 5\). The local maximum and minimum points are indeed \((5, -10)\) and \((-5, 10)\), respectively.
Step 1 ：Understand the problem: We are asked to find the local maximum and minimum values of the function \(f(x)=-x-\frac{25}{x}\) using the Second Derivative Test.
Step 2 ：Find the first derivative of the function: The first derivative of the function \(f(x)\) is \(f'(x)=-1+\frac{25}{x^2}\).
Step 3 ：Set the first derivative equal to zero and solve for \(x\): Setting \(f'(x)=0\) gives \(-1+\frac{25}{x^2}=0\). Solving for \(x\) gives \(x=\pm 5\).
Step 4 ：Find the second derivative of the function: The second derivative of the function \(f(x)\) is \(f''(x)=-\frac{50}{x^3}\).
Step 5 ：Substitute the values of \(x\) into the second derivative: Substituting \(x=5\) into \(f''(x)\) gives \(f''(5)=-2\). Substituting \(x=-5\) into \(f''(x)\) gives \(f''(-5)=2\).
Step 6 ：Apply the Second Derivative Test: The Second Derivative Test states that if \(f''(x)>0\), then \(f(x)\) has a local minimum at \(x\). If \(f''(x)<0\), then \(f(x)\) has a local maximum at \(x\).
Step 7 ：Determine the local maximum and minimum values: Since \(f''(5)<0\), \(f(x)\) has a local maximum at \(x=5\). Since \(f''(-5)>0\), \(f(x)\) has a local minimum at \(x=-5\).
Step 8 ：Find the \(y\)-coordinates of the local maximum and minimum points: Substituting \(x=5\) into \(f(x)\) gives \(f(5)=-5-5=-10\). Substituting \(x=-5\) into \(f(x)\) gives \(f(-5)=5+5=10\).
Step 9 ：\(\boxed{\text{The local maximum point is }(5, -10)\text{ and the local minimum point is }(-5, 10)}\)
Step 10 ：Check the answer: The function \(f(x)\) is continuous and differentiable for all \(x\neq 0\). The Second Derivative Test is valid for \(x=\pm 5\). The local maximum and minimum points are indeed \((5, -10)\) and \((-5, 10)\), respectively.
