Calulus derivatives

Question

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

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

(b) Find the interval(s) where  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)

To check that f(x) is increasing on its domain, we will verify that f'(x)>0 on this domain.

f'(x) = 4* cos x + 2* x + 5

Since -1<= cos x <= 1, we have that -4<= 4* cos x <= 4

Since 0<= x <= 2pi, we have that 0<=. 2* x <= 4pi

Summing the inequalities above and adding 5 to get f'(x), we get:

1<= f'(x) <= 9+ 4pi

Thus f'(x)>0, and thus f(x) is increasing on [0, 2pi].

(b)

We know that f is concave up if f''(x)>0 and concave down if f''(x)<0.

f''(x) = -4*sin x + 2

We find the inflection points by solving for x such that f''(x) = 0

-4*sinx + 2 = 0 ---> sin x = 0.5, which has the solutions x₁ = pi/6 and x₂ = 5pi/6 in the interval [0, 2pi].

f is concave up if f"(x) >0 if sin x < 0.5, which happens for x in [0, pi/6) and also for x in (5pi/6, 2pi]

f is concave down if f''(x)<0 if sin x > 0.5, which happens for x in (pi/6, 5pi/6)