5. Graphical Applications of Derivatives

Concavity

5. Graphical Applications of Derivatives

Concavity

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Multiple Choice

Determine the intervals for which the function is concave up or concave down. State the inflection points.
$f(x)=2\sin x+3x$ ; $0$ < $x$ < $2\pi$

Concave down: $\left(0,\pi\right)$ ; concave up: $\left(\pi,2\pi\right)$ ; Inflection pt: $\left(\pi,3\pi\right)$

Concave down: $\left(0,\pi\right)$ ; concave up: $\left(\pi,2\pi\right)$ ; Inflection pt: $\left(\pi,2+3\pi\right)$

Concave down: $\left(0,\frac{\pi}{2}\right)$ , $\left(\pi,\frac{3\pi}{2}\right)$ ; concave up: $\left(\frac{\pi}{2},\pi\right)$ , $\left(\frac{3\pi}{2},2\pi\right)$ ; Inflection pts: $\left(\frac{\pi}{2},\frac{4+3\pi}{2}\right)$ , $\left(\pi,3\pi\right)$ , $\left(\frac{3\pi}{2},\frac{-4+9\pi}{2}\right)$

Concave up: $\left(0,2\pi\right)$ ; No inflection pt

Verified step by step guidance

To determine concavity, we need to find the second derivative of the function f(x) = 2sin(x) + 3x. Start by finding the first derivative f'(x).

The first derivative f'(x) is obtained by differentiating each term separately: the derivative of 2sin(x) is 2cos(x), and the derivative of 3x is 3. So, f'(x) = 2cos(x) + 3.

Next, find the second derivative f''(x) by differentiating f'(x). The derivative of 2cos(x) is -2sin(x), and the derivative of 3 is 0. Therefore, f''(x) = -2sin(x).

Determine where f''(x) changes sign to find the intervals of concavity. Set f''(x) = 0 to find potential inflection points: -2sin(x) = 0, which implies sin(x) = 0. Solve for x in the interval (0, 2π).

The solutions to sin(x) = 0 within the interval (0, 2π) are x = π and x = 2π. Test intervals around these points to determine where f''(x) is positive (concave up) or negative (concave down). The function is concave down on (0, π) and concave up on (π, 2π). The inflection point is at x = π.