Question

Concavity Determine the intervals on which the following functions are concave up or concave down. Identify any inflection points. $$h(t)=2+\cos 2 t \text { on }[0, \pi]$$

Short answer

Question: Determine the intervals where the function $$h(t)=2+\cos 2t$$ is concave up and concave down, as well as the inflection points, on the interval [0, π]. Answer: The function is concave up on the interval $$\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$$ and concave down on the intervals $$\left(0, \frac{\pi}{4}\right) \cup \left(\frac{3\pi}{4}, \pi\right)$$. The inflection points are at $$t=\frac{\pi}{4}$$ and $$t=\frac{3\pi}{4}$$.

Step-by-step solution

Find the first and second derivatives.

To find the concavity and inflection points, we need to determine the second derivative of the function. First, we'll find the first derivative and then the second derivative of the function. Function: $$h(t)=2+\cos 2t$$ First derivative: $$h'(t)=-2\sin(2t)$$ Second derivative: $$h''(t)=-4\cos(2t)$$

Find the critical points of the second derivative.

To find the critical points, we need to determine where the second derivative is equal to zero or undefined. The second derivative is continuous and defined for all values, so we only need to find where it is equal to zero. $$-4\cos(2t)=0$$ Divide both sides by -4: $$\cos(2t)=0$$ We know that cosine is zero at odd multiples of $$\frac{\pi}{2}$$, so we must find the critical points in the range of [0, π]: $$2t=\frac{\pi}{2},\frac{3\pi}{2}$$ Divide both sides by 2 to find \(t\): $$t = \frac{\pi}{4}, \frac{3\pi}{4}$$

Analyze the concavity.

Evaluate the second derivative on either side of the critical points $$\frac{\pi}{4}$$ and $$\frac{3\pi}{4}$$ to determine the concavity of the function. Test interval: $$0 < t < \frac{\pi}{4}$$, let's use $$t=\frac{\pi}{8}$$. $$h''\left(\frac{\pi}{8}\right)=-4\cos\left(\frac{\pi}{4}\right)=-4\left(\frac{\sqrt{2}}{2}\right)<0$$ The function is concave down on this interval. Test interval: $$\frac{\pi}{4} < t < \frac{3\pi}{4}$$, let's use $$t=\frac{\pi}{2}$$. $$h''\left(\frac{\pi}{2}\right)=-4\cos(\pi)=4>0$$ The function is concave up on this interval. Test interval: $$\frac{3\pi}{4} < t < \pi$$, let's use $$t=\frac{7\pi}{8}$$. $$h''\left(\frac{7\pi}{8}\right)=-4\cos\left(\frac{7\pi}{4}\right)=-4\left(\frac{\sqrt{2}}{2}\right)<0$$ The function is concave down on this interval.

Identify the inflection points and intervals of concavity.

The inflection points occur when the concavity changes, which happens at $$t=\frac{\pi}{4}$$ and $$t=\frac{3\pi}{4}$$. Inflection points: $$IP_1 = \left(\frac{\pi}{4}, h\left(\frac{\pi}{4}\right)\right)$$ $$IP_2 = \left(\frac{3\pi}{4}, h\left(\frac{3\pi}{4}\right)\right)$$ Intervals of concavity: Concave up: $$\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)$$ Concave down: $$\left(0, \frac{\pi}{4}\right) \cup \left(\frac{3\pi}{4}, \pi\right)$$

Key concepts

Second Derivative

The second derivative is a powerful tool in calculus, particularly when examining the concavity of functions. To understand concavity, we first need to find the second derivative of the original function. This derivative gives us information about the rate at which the slope of the function is changing. If a function’s first derivative is denoted as \( f'(x) \), then the second derivative will be \( f''(x) \). Essentially, the second derivative tells us how the rate of change of the function itself is changing.

The second derivative test is used to determine the concavity of a function:

• If \( f''(x) > 0 \) for an interval, the function is concave up on that interval.
• If \( f''(x) < 0 \) for an interval, the function is concave down on that interval.

In our case, we have the function \( h(t) = 2 + \cos(2t) \), with its second derivative \( h''(t) = -4\cos(2t) \). By assessing where this second derivative is positive or negative, we can identify intervals of concavity.

Inflection Points

Inflection points are points on the graph of a function where the concavity changes from up to down or from down to up. To find inflection points, you need to look for values of \( x \) where the second derivative either equals zero or is undefined. However, just finding where \( f''(x) \) equals zero is not enough; we must also ensure that the concavity actually changes at those points.

In practical terms, this means checking the sign of the second derivative on both sides of the point:

• If the sign of \( f''(x) \) changes from positive to negative or negative to positive as you pass through \( x \), then \( x \) is an inflection point.

For our function \( h(t) \), the second derivative \( h''(t) \) is zero at \( t = \frac{\pi}{4} \) and \( t = \frac{3\pi}{4} \). At these points, the concavity changes, making them inflection points. Thus, the inflection points for \( h(t) \) are \( \left(\frac{\pi}{4}, h\left(\frac{\pi}{4}\right)\right) \) and \( \left(\frac{3\pi}{4}, h\left(\frac{3\pi}{4}\right)\right) \).

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in the analysis of periodic behaviors in mathematics. These functions are often involved in calculus problems due to their oscillatory nature, which is particularly interesting when studying derivatives and concavity. For instance, the function \( h(t) = 2 + \cos(2t) \) involves the cosine function.

The behavior of trigonometric functions is essential when evaluating the second derivative:

• The cosine function has a period of \( 2\pi \), meaning it repeats its values every \( 2\pi \) units.
• Key values occur at even or odd multiples of \( \frac{\pi}{2} \), such as \( t = \frac{\pi}{4} \) or \( t = \frac{3\pi}{4} \) in the given range.

By understanding these properties, you can determine critical points and points of concavity in functions involving trigonometric components. For trigonometric functions, recognizing these key values helps in finding zeros of the second derivative, as shown in previous analysis of \( -4\cos(2t) \).