Calculus Questions with Answers (1)

Calculus Questions with Answers (1)

Calculus questions with detailed solutions are presented. The uses of the first and second derivative to determine the intervals of increase and decrease of a function, the maximum and minimum points, the interval(s) of concavity and points of inflections are discussed.

Question 1

For the function f(x) = (x + 3)(x - 2) 3 find
a) the interval(s) of increase and decrease of f,
b) the value(s) of x for which f has a local maximum and minimum,
c) and the interval of concavity and inflection point(s).

Solution to Question 1:

  • a) We first find the first derivative f '
    f '(x) = (x - 2) 3 + 3(x + 3)(x - 2) 2
    = (x - 2) 2(4x + 7)
  • f'(x) has zeros at
    are x = 2 and x = - 7/4
  • The table of sign of f '(x) is shown below
    table of sign of first derivative f', question 1

  • Using the table, f is decreasing on the interval
    (-infinity , - 7/3)
  • and increasing on the interval
    (- 7/3 , +infinity)
  • b) f has a local minimum at x = - 7/3. Note that although f '(2) = 0, there is no local extremum at x = 2. This is because f ' (x) does not change sign at x = 2.
  • c) We now calculate f"(x)
    f"(x) = 2(x - 2)(4x + 7) + 4(x - 2) 2
    = 6(x - 2)(2x + 1)
  • The sign table of f"(x)is shown below
    table of sign of second derivative f

  • Using the above table, we can say that the graph of f is concave up on the intervals (-infinity , -1/2) and (2 , +infinity) and concave down on the interval (-1/2 , 2).


Question 2

Given f(x) = e (x 2 - x), find
a) the interval(s) of increase and decrease of f,
b) value(s) of x for which f has a local extremum,
c) the interval(s) of concavity and any inflection point(s).

Solution to Question 2:

  • Find f '(x)
    a) f '(x) = (2x - 1) e (x 2 - x)
  • f '(x) has one zero at x = 1/2. The sign table of f '(x) is shown below
    table of sign of first derivative f ', question 2

  • f is decreasing on (-infinity , + 1/2) and increasing on (1/2 , +infinity).
  • b) Since x = 1/2 is a critical value and f ' changes sign at x = 1/2, there is a local minimum at x = 1/2.
  • c) We first calculate the second derivative f" of f.
    f"(x) = 2 e (x 2 - x) + (2x-1)(2x-1)e (x 2 - x)
    = [ 2 + (2x+1)2 ] e (x 2 - x)
  • The second derivative f" is always positive. Hence the graph of f is concave up on (-infinity , +infinity) and since f" does not change sign, the graph of f has no point of inflection.

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