Question

You use 'INF' for [infinity][infinity] and '-INF' for −[infinity]−[infinity].

And use 'U' for the union symbol.
Enter DNE if an answer does not exist.

f(x)=4sinx+3cosx,on(−π,π)f(x)=4sin⁡x+3cos⁡x,on(−π,π)

a) Determine the intervals on which ff is concave up and concave down.
ff is concave up on:
ff is concave down on:

Answer 1

To find intervals of concavity for the function f(x) = 4sin(x) + 3cos(x), we calculate its second derivative and analyze its sign changes. Critical points are found by setting the second derivative to zero.

Step-by-step explanation:

To determine the intervals on which the function f(x) = 4sin(x) + 3cos(x) is concave up or concave down on the interval −π, π, we must examine the function's second derivative.

First, we find the first derivative f'(x), which is 4cos(x) - 3sin(x). Then we calculate the second derivative f''(x), which is -4sin(x) - 3cos(x).

For the function to be concave up, f''(x) must be greater than zero, and for it to be concave down, f''(x) must be less than zero. To find where f''(x) is greater or less than zero, we set f''(x) equal to zero and solve for x. This gives us the critical points that potentially separate intervals of concavity.

We then test intervals around these critical points to determine the sign of f''(x) and hence the concavity of f(x).

Let's take an example to better illustrate this process. Suppose we found that f''(x) = 0 at x = -2 and x = 1 within the domain −π, π.

We would then test values to the left of x = -2, between x = -2 and x = 1, and to the right of x = 1 to determine the concavity in those intervals. If f''(x) is positive on an interval, the function is concave up there, and if it's negative, the function is concave down.