Question

Prove or disprove:
At x=2, the function 3x3+12x2- 48 x + 25 attains its maximum value on the interval [0,3].​

Answer 1

we have to check, at x = 2, the function 3x³ + 12x² - 48x + 25 attains its maximum value on the interval [0, 3].

solution : function, f(x) = 3x³ + 12x² - 48x + 25

differentiating f(x) with respect to x,

f'(x) = 9x² + 24x - 48

at f'(x) = 0, 9x² + 24x - 48 = 0

⇒3x² + 8x - 16 = 0

⇒3x² + 12x - 4x - 16 = 0

⇒3x(x + 4) - 4(x + 4) = 0

⇒(3x - 4)(x + 4) = 0

⇒x = 4/3 , -4

now differentiating once again with respect to x,

f"(x) = 18x + 24

at x = 4/3 , f"(x) > 0 so at x = 4/3, f is minimum.

at x = -4, f"(x) < 0, so at x = -4 , f is maximum.

Therefore it is clear that, at x = 2 f attains neither maximum nor minimum. so given statement is false. we disapproved it.

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Answer 2

Answer:

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Step-by-step explanation:

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