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15 September, 23:41

In exercise, find the relative extrema of the function.

f (x) = x4 - 2x2 + 5

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Answers (1)
  1. 16 September, 02:45
    0
    minima at x = ±1

    Step-by-step explanation:

    Given function in the question:

    f (x) = x⁴ - 2x² + 5

    now,

    To find the extrema differentiating the function with respect to 'x' and equate it to zero

    we get

    f' (x) = 4x³ - (2) (2x) + 0 = 0

    or

    4x³ - (2) (2x) = 0

    or

    4x³ - 4x = 0

    or

    x³ - x = 0

    or

    x (x² - 1) = 0

    or

    x = 0 and x = ±1

    checking for maxima or minima

    again differentiating f' (x) with respect to x, we get

    f'' (x) = (3) 4x² - x

    or

    f'' (x) = 12x² - x

    at x = 0

    f" (0) = 12 (0) ² - 0 = 0 [neither maxima nor minima]

    at x = - 1

    f" (-1) = 12 (-1) ² - (-1) = 12 + 1 = 13 > 0 [relative minima]

    at x = 1

    f" (1) = 12 (1) ² - 1 = 12 - 1 = 11 > 0 [relative minima]

    hence,

    minima at x = ±1
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