Application of Derivatives - II

  • We can apply derivatives to determine approximate value of certain quantities and to find maximum and minimum values of a function in an interval.
  • The differential of y is denoted by dy and is defined by dy = f ′(x) dx or dy = f ′(x) ∆x, where dy approximately equal to ∆y.
  • First Derivative Test for Maxima and Minima:

Let f (x) be a function defined on an open interval I and continuous at a critical point ‘a’ in I. Then

  • If f ′(x) changes sign from positive to negative as x increases through a, i.e.,

if f ′(x) > 0 at every point sufficiently close to and to the left of a, and f ′(x) < 0 at every point sufficiently close to and to the right of a, then a is a point of local maxima.

  • If f′(x) changes sign from negative to positive as x increases through a, i.e., if f′(x) < 0 at every point sufficiently close to and to the left of a, and f ′(x) > 0 at every point sufficiently close to and to the right of a, then a is a point of local minima.
  • Second Derivative Test  For Maxima and Minima:
  • Let f(x) be a function defined on an interval I and a Î I. Let f(x) be twice differentiable at a. Then
  • x = a is a point of local maxima, and if f ′(a) = 0 and f ″(a) < 0, then f(a) is the local maximum value of f (x).
  • x = a is a point of local minima, and if f′(a) = 0 and f ″(a) > 0, the f (a) is the local minimum value of f (x).
  • The test fails if f'(a) = 0 and f" (a) = 0.

       Keywords: Maximum and minimum values of a function, First derivative test, Second derivative test

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  • Q1

    Marks:1
    Answer:

    2.

    Explanation:

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  • Q2

    The maximum and minimum values of 4 cos x + 3 sin x are respectively

    Marks:1
    Answer:

    5 and -5.

    Explanation:
    The maximum and minimum values of a cos x + b sin x are (a2+b2) and -(a2+b2) respectively.
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  • Q3

    The approximate change in the volume V of a cube of side a metres caused by increasing the side by 5% is (in m3)

    Marks:1
    Answer:

    0.15 a3.

    Explanation:
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  • Q4

    If f(x) = 3x2 + 5x + 3, then the approximate value of f(3.02) is

    Marks:1
    Answer:

    45.46.

    Explanation:
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  • Q5

    Approximate value of (0.99) is

    Marks:1
    Answer:

    0.995

    Explanation:

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