Question

Determine the open intervals on which the graph of the function is concave upward or concave downward.

y=7 x+5/sin (x), (-π, π)
find, concave upward .

Answer 1

The graph of the function y = 7x + 5/sin(x) is concave upward on the intervals (-π/2, π/6)U(π/2, 5π/6)U(9π/6, 3π/2).

Step-by-step explanation:

The graph of the function y = 7x + 5/sin(x) can be concave upward or concave downward on different intervals. To determine where the graph is concave upward or concave downward, we need to find the second derivative of the function and determine its sign. Let's start by finding the first and second derivatives:

First derivative: y' = 7 - (5cos(x))/sin^2(x)

Second derivative: y'' = (5cos(x)(3sin^2(x) - 2))/(sin^3(x))

To determine the intervals where the graph is concave upward or concave downward, we need to find where the second derivative is positive or negative. In this case, the second derivative is positive when (3sin^2(x) - 2)cos(x) > 0. We know that cos(x) is positive on the interval (-π/2, π/2) where sin(x) > 0. Therefore, the second derivative is positive when (3sin^2(x) - 2) > 0, which means sin(x) > sqrt(2/3) or sin(x) < -sqrt(2/3). This occurs on the intervals (-π/2, π/6)U(π/2, 5π/6)U(9π/6, 3π/2).

Therefore, the graph of the function y = 7x + 5/sin(x) is concave upward on the intervals (-π/2, π/6)U(π/2, 5π/6)U(9π/6, 3π/2).