The graph of derivative f ′(x) of some function f (x) is given below.

Question

(a) Identify the subinterval(s) of (-4,3) where f (x) has a constant slope. Explain why?

(b) Identify the subinterval(s) of (-4,3) where f (x) is decreasing. Explain why?

(c) Identify the x-coordinate(s) where f (x) attains a local maximum value, local minimum value (if any). Explain why?

(d) Identify the subinterval(s) of (-4,3) where f (x) is concave down. Explain why?

Answer 1

a) x ∈ (-4 , -1) in other words -4 < x < -1. Since f^' gives the slope of f and f^' is constant on that interval.

b) f is decreasing on (-1 , -.7) U (1 , 2.7) since f^' < 0 there (the graph of f^' is below the x-axis).

c) f has relative minima at x = -.7 and x = 2.7 since f^' changes from negative to positive there, and f has a relative maximum at x = 1, since f^' goes from positive to negative there.

d) f is concave down on (0 , 2), since f^' is decreasing on that interval.

Answer 2

Actually, I manage to get the functions and expressions and I use Desmos to see the graph clearly. Here are the piecewise function for f ' (x) and f(x).

f ' (x) = { 3 if -4<x<-1

x³ - 3x² + 2 if -1<x<3

From there, I got the f(x) by getting the antiderivative of f ' (x):

f(x) = { 3x + C if -4<x<-1

x⁴/4 - x³ + 2x + C if -1<x<3

In this case, you have to change the way you think because the graph is f ' (x). Here is my explanation:

a. The f(x) has a constant slope at the subinterval of x (-4,-1) because that part of the graph f ' (x) is a straight horizontal line segment.

b. The f(x) is decreasing at the subinterval of x (-1, -0.732) and (1,2.732) because that part of the graph f ' (x) is below the x-axis.

c. Yes, f(x) attains local minima at x = -7.32 and x = 2.732, and a local maximum at x=1. We can think of the graph of f ' (x) as a third degree polynomial with positive leading coefficient because of its end behavior and 2 extrema. If that's the case, f(x) is a fourth degree polynomial function with positive leading coefficient and it has at most 3 local extrema at f ' (x) = 0. The one in the middle is the local maximum and the other two are the local minima.

d.

At subinterval (-1,0), f(x) is open up because f ' (x) goes up.

At subinterval (0,2), f(x) is open down because f'(x) goes down.

At subintercal (2,3), f(x) is open up because f ' (x) goes up.