Question

In Exercises, find all relative extrema of the function. Use the Second- Derivative Test when applicable. $$ f(x)=\sqrt{2 x^{2}+6} $$

Short answer

The function \( f(x) = \sqrt{2x^2 + 6} \) has a local minimum at \( x = 0 \).

Step-by-step solution

Compute the First Derivative

Find the first derivative of \( f \). This can be done by the chain rule for differentiation as follows: \( f'(x) = \frac{1}{2}(2x^2+6)^{-1/2}.2 \cdot 2x = \frac{2x}{\sqrt{2x^2+6}} \).

Find Extrema Candidates

Set the first derivative equal to zero and solve for \( x \). We have \( f'(x) = 0 => \frac{2x}{\sqrt{2x^2+6}} = 0 \). From this equation, we find \( x = 0 \) as the potential extremum point.

Compute the Second Derivative

Find the second derivative of \( f \). This can also be done by the chain rule and product rule together: \( f''(x) = \frac{\sqrt{2x^2+6}\cdot 2 - 2x \cdot \frac{x}{\sqrt{2x^2+6}}}{2x^2+6} = \frac{2}{\sqrt{2x^2+6}} - \frac{2x^2}{2x^2+6} \).

Evaluate the Second Derivative at the Extrema Candidates

At \( x = 0 \), we find \( f''(0) = 2/\sqrt{6} > 0 \). According to the Second-Derivative Test, if \( f''(x) > 0 \) at a point, then that point is a local minimum.

Conclusion

The function \( f(x) = \sqrt{2x^2 + 6} \) has a local minimum at \( x = 0 \).

Key concepts

Second-Derivative Test

The Second-Derivative Test is a useful tool in calculus to determine if a point is a local extremum, which could either be a local minimum or a local maximum. After finding the first derivative and identifying points where it is zero (these are your candidates for extrema), you can use the second derivative to test these points more thoroughly.

• If the second derivative, denoted as \( f''(x) \), is positive at a candidate point, this indicates that the function is concave up at that point and generally means you have a local minimum.
• If \( f''(x) \) is negative, the function is concave down, pointing to a local maximum.
• If \( f''(x) \) equals zero, the test is inconclusive, and you might need to investigate further, possibly with other methods.

In our problem, by evaluating the second derivative at \( x = 0 \), we see that \( f''(0) > 0 \), leading us to conclude there is a local minimum at this point.

First Derivative

The first derivative of a function, often denoted as \( f'(x) \), shows how the function's output changes with respect to changes in the input \( x \). This derivative provides information about the slope or the rate of change of the function at any given point.

• When \( f'(x) > 0 \), the function is increasing at that interval.
• If \( f'(x) < 0 \), then the function is decreasing.
• A critical point occurs where \( f'(x) = 0 \), which might suggest a local extremum (either a minimum or maximum).

In the problem at hand, finding \( f'(x) = \frac{2x}{\sqrt{2x^2+6}} = 0 \) enabled us to locate \( x = 0 \) as a potential point for extrema. This crucial step points us in the direction of further analysis with the second derivative.

Chain Rule

The Chain Rule is essential when dealing with derivatives of composite functions. It allows us to differentiate functions in a step-by-step manner by breaking them into their constituent parts. Consider a composite function \( f(g(x)) \), the chain rule states:

• First, take the derivative of the outer function \( f \) with respect to \( g \), treating \( g(x) \) as the argument.
• Then multiply by the derivative of the inner function \( g(x) \) with respect to \( x \).

In our example, we apply the chain rule when differentiating \( f(x)=\sqrt{2x^2+6} \). Here, the inner function is \( g(x)=2x^2+6 \) with the outer function being \( f(g(x))=\sqrt{g(x)} \). Thus:\[ f'(x) = \frac{1}{2}(2x^2+6)^{-1/2} \cdot (4x) = \frac{2x}{\sqrt{2x^2+6}}.\]This application of the chain rule lets us find how rapidly the original function changes, which is central in determining extrema locations.