Problem 63 - Calculus Early Transcendentals

Question

Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$g(t)=\ln \left(3 t^{2}+1\right)$$

Step-by-Step Solution

Answer

Question: At which intervals is the function \(g(t)=\ln \left(3 t^{2}+1\right)\) concave up, concave down, and identify the inflection points. Answer: The function is concave up in the intervals \((-\infty, -\frac{1}{\sqrt{3}})\) and \((-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})\), concave down in the interval \((\frac{1}{\sqrt{3}}, \infty)\), and has inflection points at \(t_1 = \frac{1}{\sqrt{3}}\) and \(t_2 = -\frac{1}{\sqrt{3}}\).

Step 1: Find the first derivative of the function

To find the first derivative, we use the chain rule on the given function: $$g(t)=\ln \left(3 t^{2}+1\right)$$ $$g'(t) = \frac{1}{3t^2 + 1} \cdot \frac{d(3t^2 + 1)}{dt}$$ Now, find the derivative of the inner function: $$\frac{d(3t^2 + 1)}{dt} = 6t$$ So the first derivative is: $$g'(t) = \frac{6t}{3t^2 + 1}$$

Step 2: Find the second derivative of the function

Now, we will find the second derivative by differentiating the first derivative we found: $$g''(t) = \frac{d}{dt}\left(\frac{6t}{3t^2 + 1}\right)$$ To find the second derivative, we will use the quotient rule: $$g''(t) = \frac{(3t^2 + 1)(6) - 6t(6t)}{(3t^2 + 1)^2}$$ Simplify the second derivative: $$g''(t) = \frac{18t^2 + 6 - 36t^2}{(3t^2 + 1)^2}$$ $$g''(t) = \frac{-18t^2 + 6}{(3t^2 + 1)^2}$$

Step 3: Find the critical points

A critical point is a point where the second derivative is either equal to zero or undefined. Set the second derivative equal to zero and solve for t: $$-\frac{18t^2 + 6}{(3t^2 + 1)^2} = 0$$ $$\Rightarrow -18t^2 + 6 = 0$$ $$\Rightarrow 6 = 18t^2$$ $$\Rightarrow t^2 = \frac{1}{3}$$ $$\Rightarrow t = \pm\frac{1}{\sqrt{3}}$$ Now, we have two critical points: \(t_1 = \frac{1}{\sqrt{3}}\) and \(t_2 = -\frac{1}{\sqrt{3}}\)

Step 4: Test the intervals

Now, we will test the intervals between the critical points to determine concavity: 1. Test the interval \(t < -\frac{1}{\sqrt{3}}\): Choose \(t = -1\), so: $$g''(-1) = \frac{-18(-1)^2 + 6}{(3(-1)^2 + 1)^2} > 0$$ Since \(g''(-1) > 0\), the function is concave up in this interval. 2. Test the interval \(-\frac{1}{\sqrt{3}} < t < \frac{1}{\sqrt{3}}\): Choose \(t = 0\), so: $$g''(0) = \frac{-18(0)^2 + 6}{(3(0)^2 + 1)^2} > 0$$ Since \(g''(0) > 0\), the function is concave up in this interval. 3. Test the interval \(t > \frac{1}{\sqrt{3}}\): Choose \(t = 1\), so: $$g''(1) = \frac{-18(1)^2 + 6}{(3(1)^2 + 1)^2} < 0$$ Since \(g''(1) < 0\), the function is concave down in this interval.

Step 5: Identify inflection points and concave intervals

Inflection points occur at the critical points \(t_1 = \frac{1}{\sqrt{3}}\) and \(t_2 = -\frac{1}{\sqrt{3}}\) The function is concave up in the intervals: \((-\infty, -\frac{1}{\sqrt{3}})\) and \((-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})\) The function is concave down in the interval: \((\frac{1}{\sqrt{3}}, \infty)\)

Key Concepts

Concave Up and Down

Understanding the concavity of functions is crucial for visualizing how a graph curves and predicting its behavior. A function is concave up on an interval if its graph bends upwards, creating a 'u' shape. Here, the second derivative of the function is positive. This indicates that as you move from left to right, the slope of the tangent line to the function is increasing.

Conversely, a function is concave down when its graph forms a 'n' shape, curving downwards on an interval, with the second derivative being negative. This implies that the slope of the tangent is decreasing as you travel from left to right along the graph. Through concavity analysis, we can locate regions where a function is opening upwards or downwards, revealing important aspects of its geometry and potential applications.

Inflection Points

Inflection points are points on the graph of a function where the concavity changes from up to down or vice versa. These points are especially interesting because they mark the transition in the curvature of the function's graph. Finding an inflection point involves looking where the second derivative of a function equals zero or is undefined, and then determining if there's an actual change in concavity.

However, not every instance where the second derivative is zero will provide an inflection point. To confirm, we must check the concavity on either side of these critical values. If the sign of the second derivative changes, then it's indeed an inflection point. These points can sometimes indicate relative maxima or minima, making them valuable for optimization problems.

Second Derivative Test

The second derivative test is a handy analytical tool in calculus to determine the concavity of a function and ascertain whether a critical point is a local maximum, minimum, or neither. After locating the critical points by setting the first derivative equal to zero, we apply this test by assessing the sign of the second derivative at these points.

If the second derivative is positive at a critical point, it is a local minimum. If it's negative, we have a local maximum. And if the second derivative is zero or undefined, the test is inconclusive; we would have to use other methods, such as the first derivative test, to analyze the function's behavior near such points. This test is particularly advantageous when dealing with functions where the graph isn't easily sketched.

Chain Rule

Differentiating complex functions often requires the chain rule, one of the fundamental tools in calculus. The rule is used when we have a 'function within a function', technically called a composite function. To apply the chain rule, we differentiate the outer function as if the inner function were just a variable and then multiply it by the derivative of the inner function.

The first step in our solution involves applying the chain rule. In the context of the provided exercise, we differentiate the logarithmic function treating \(3t^2 + 1\) as the inner function. Then, we multiply this result by the derivative of the inner function itself, which in this case is \(6t\), yielding the first derivative of our original function.

Quotient Rule

When we need to differentiate a function that is a quotient of two other functions, we turn to the quotient rule. This rule states that the derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all over the square of the denominator. Symbolically, if \(h(x) = \frac{f(x)}{g(x)}\), then \(h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}\).

The quotient rule is particularly useful when the numerator and denominator of a fraction cannot be simplified or separated into individual terms that can be differentiated on their own. In our example, to find the second derivative, we applied the quotient rule to the first derivative, selecting convenient points to test the sign of the second derivative and determine the function's concavity.