The picture below shows the graph y=f′(x) y = f ′ ( x ) of the derivative of a function y=f(x) y = f ( x ) .

Question

The picture below shows the graph y=f′(x)of the derivative of a function y=f(x)

For each of the labeled points on the graph, classify the corresponding point on the graph of y=f(x) as one of the following: MAX, MIN, INFL, INT (short for maximum, minimum, inflection point, x-intercept) 

A:___

B:___

C:___

D:___

E:___

For each of the following intervals, classify whether the graph of y=f(x) is INC or DEC over that interval (short for increasing or decreasing). 

(-∞, A) ___

(A, B) ___

(B, C) ___

(C, D) ___

(D, E) ___

(E, ∞) ___

For each of the following intervals, classify whether the graph of y=f(x) is CU or CD over that interval (short for concave up or concave down). 

(-∞, A) ___

(A, B) ___

(B, C) ___

(C, D) ___

(D, E) ___

(E. ∞) ___

Answer 1

Key points:

* f’(x) yields the slope of f(x) at a given point. Therefore, wherever, f’(x) > 0, f(x) is increasing. Wherever f’(x) < 0, f(x) is decreasing.

* Points at which f’(x) = 0 are critical points (MAX, MIN, or INFL). Let x₀ be such a point. If f(x₀) is increasing left of x₀ and decreasing right of x₀ then x₀ is a maximum. If f(x₀) is increasing left of x₀ and decreasing right of x₀ then x₀ is a maximum. Conversely, if f(x₀) is decreasing left of x₀ and increasing right of x₀ then x₀ is a minimum.

* f’’(x) represents the slope of f’(x). Points at which f’’(x) = 0 are potential inflection points. Let x₀ be such a point. If the concavity of f(x₀) changes at x₀, then x₀ is an inflection point.

* In a given interval, if f’’(x) > 0, f(x) is concave upward. Conversely, if f’’(x) < 0, f(x) is concave downward. Thus, f(x) is concave upward when the slope of f’(x) > 0, and concave downward when the slope of f’(x) > 0.

Using these pointers:

A: MIN; B: INFL; C: MAX; D: INFL; E: MIN

(-∞,A): DEC; (A,B): INC; (B,C): INC; (C,D): DEC; (D,E): DEC

(-∞,A): CU; (A,B): CU; (B,C): CD; (C,D): CD; (D,E): CU