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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