Calculus concavity

Question

Consider the function  𝑓(𝑥)=4sin𝑥+𝑥2+5𝑥−10 with domain the closed interval [0,2𝜋]

(a) Show that  𝑓(𝑥)f(x) is increasing on its domain. Explain clearly your reasoning.

(b) Find the interval(s) where  𝑓(𝑥)f(x) is concave up and the interval(s) where it is concave down, and any points of inflection. Show your work.

Answer 1

A function is increasing when the first derivative is positive.

For f(x) = 4sin𝑥 + 𝑥² + 5𝑥 − 10, f '(x) = 4cos(x) + 2x + 5

Since cos(x) is a function that oscillates between -1 and +1, the smallest value would be -1. Since the domain is positive or zero, "2x" will be positive or zero. We are adding 5 to the result therefore the result must be > 4 on the domain, therefore the function is increasing.

The 2nd derivative tells if the function is concave up or down (concave up when f ''(x) > 0 and concave down when f ''(x) < 0). Since f '(x) = 4cos(x) + 2x + 5, then f ''(x) = -4sin(x) + 2

To find where f '' is > 0 and < 0, lets find where it equals zero. This will be a boundary. So:

-4sin(x) + 2 = 0

4sin(x) = 2

sin(x) = 1/2

This occurs at x = π/6 and x = 5π/6

So create a number line to consider all values in the domain separated at these boundaries:

You can try a number in each of the sections in the f '' function to see if you get a positive or a negative. If you get positive, the function is concave up in this interval and if you get negative, its CC down.

For the first interval, try π/12

f ''(π/12) = -4sin(π/12) + 2 = 0.965 and since it is positive, the function is CC Up on (0, π/6)

For the 2nd interval, try π/2

f ''(π/2) = -4sin(π/2) + 2 = -2 and since it is negative, the function is CC Down on (π/6, 5π/6)

For the 3rd interval, try π

f ''(π) = -4sin(π) + 2 = 2 and since it is positive, the function is CC Up on (5π/6, 2π)

Points of inflection occur when f '' = 0 and when the concavity changes there. So the POI at x = π/6 and x = 5π/6