Problem 26 - Essential Calculus

Question

Find all relative extrema. Use the Second Derivative Test where applicable. \(f(x)=\sqrt{x^{2}+1}\)

Step-by-Step Solution

Answer

The function \(f(x)=\sqrt{x^{2}+1}\) has one relative minimum at \(x=0\).

Step 1: Differentiation

Firstly, we need to calculate the derivative of the function. A useful formula for this task is the chain rule. For a function given \(h(u)=\sqrt{u}\), its derivative is \(\frac{1}{2\sqrt{u}}\cdot u'\), so we have \(f'(x)=\frac{x}{\sqrt{x^{2}+1}}\).

Step 2: Find the Critical Points

The critical points are the values of the variable where the function has a local maximum or minimum. These critical points can be found by setting the derivative equal to zero and solve for \(x\), which gives \(\frac{x}{\sqrt{x^{2}+1}}=0\), and hence, \(x = 0\) is the only critical point of the function.

Step 3: Second Derivative

Next, we calculate the second derivative. Calculating the second derivative, \(f''(x)\), is a bit complex and requires use of the quotient rule and chain rule again. After some simplification, we get \(f''(x)=\frac{1}{\left(x^{2}+1\right)^{\frac{3}{2}}}\).

Step 4: Apply the Second Derivative Test

The Second Derivative Test involves substituting the critical points into the second derivative. If the value we get is positive, the function has a relative minimum at that point, if it's negative, a relative maximum. For the point \(x=0\), we have \(f''(0) = 1\), which is greater than zero.

Step 5: Identify and Label Extrema

Since the second derivative at \(x=0\) is positive, the function \(f(x)\) has a relative minimum at \(x=0\) by the Second Derivative Test.