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Find the value of x for which the function f(x) = 2x3 - x2 - 4x + 4 has a maximu...


Question

Find the value of x for which the function f(x) = 2x3 - x2 - 4x + 4 has a maximum value

Options

A) \(\frac{2}{3}\)

B) 1

C) -1

D) -\(\frac{2}{3}\)

The correct answer is B.

Explanation:

f(x) = 2x3 - x2 - 4x + 4
f(x) = 6x2 - 2x - 4 at turning point, f1(x) = 0
6x2 - 2x - 4 = 0, 3x2 - x - 2 = 0, 3x2 - 3x + 2x - 2 = 0
(3x + 2)(x - 1) = 0, x = -\(\frac{2}{3}\) or 1
f11(x) = 12x - 2,
when x = \(\frac{2}{3}\), f11(x) = 12(-\(\frac{2}{3}\)) - 2 = -10 < 0

\(\to\) f(x) is maximum @ x = -\(\frac{2}{3}\)
when x = 1, f11(x) = 12(1)- 2 = 10 > 0
\(\to\) f(x) is maximum @ x = 1


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Dicussion (1)

  • f(x) = 2x3 - x2 - 4x + 4
    f(x) = 6x2 - 2x - 4 at turning point, f1(x) = 0
    6x2 - 2x - 4 = 0, 3x2 - x - 2 = 0, 3x2 - 3x + 2x - 2 = 0
    (3x + 2)(x - 1) = 0, x = -\(\frac{2}{3}\) or 1
    f11(x) = 12x - 2,
    when x = \(\frac{2}{3}\), f11(x) = 12(-\(\frac{2}{3}\)) - 2 = -10 < 0

    \(\to\) f(x) is maximum @ x = -\(\frac{2}{3}\)
    when x = 1, f11(x) = 12(1)- 2 = 10 > 0
    \(\to\) f(x) is maximum @ x = 1

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