Thank you - (Enter your answers using interval notation. If an answer does not exist, enter DNE.)
y = 4x + [5/sin x], (−π, π)
concave upward (x,y)-
concave downward (x,y)-
y = 4x + [5/sin x]
= 4x + 5cscx
y' = 4 - 5 cscx cotx
y" = 5/2 csc^3(x) (cos(2x)+3)
cos(2x)+3 is always positive.
on (0,π) csc^3 > 0, so, y is concave up.
similarly, y is concave down on (-π,0)
To find the intervals where the graph of the equation is concave upward or concave downward, we can use the second derivative test.
First, let's find the second derivative of the equation.
Given: y = 4x + [5/sin x]
To find the second derivative, we need to differentiate the equation twice.
1. Find the first derivative:
Taking the derivative of each term, we get:
dy/dx = d/dx(4x) + d/dx(5/sin(x))
The derivative of 4x is 4, and the derivative of 5/sin(x) can be found using the quotient rule as:
= (5 * d/dx(sin(x)) - sin(x) * d/dx(5))/ sin(x)^2
= (5 cos(x) - 0)/ sin(x)^2
= 5 cos(x)/sin(x)^2
= 5cot(x)
So, dy/dx = 4 + 5cot(x)
2. Find the second derivative:
Taking the derivative of dy/dx, we get:
d^2y/dx^2 = d/dx(4 + 5cot(x))
The derivative of 4 is 0, and the derivative of 5cot(x) can be found using the chain rule as:
= -5csc^2(x)
So, d^2y/dx^2 = -5csc^2(x)
Now, let's analyze the concavity:
For the graph to be concave upward, we need d^2y/dx^2 > 0.
-5csc^2(x) > 0
Cosecant function is positive in the intervals where sin(x) < 0 (in the graph of sin(x), below the x-axis).
In the interval from -π to π, sin(x) is negative in the interval from -π to 0.
Therefore, the interval where the graph is concave upward is (0, π).
For the graph to be concave downward, we need d^2y/dx^2 < 0.
-5csc^2(x) < 0
Cosecant function is negative in the intervals where sin(x) > 0 (in the graph of sin(x), above the x-axis).
In the interval from -π to π, sin(x) is positive in the interval from 0 to π.
Therefore, the interval where the graph is concave downward is (−π, 0).
So, the answers are:
Concave upward (x,y): (0, π)
Concave downward (x,y): (−π, 0)