Question

find the local maximum and local mimnimun value of f(x)= x ^{2} - 3x^2 - 24x + 5

​Answer:

Step by Step explanation:​

Answer 1

EXPLANATION.

Local maximum and local minimum of the value.

⇒ f(x) = x³ - 3x² - 24x + 5.

As we know that,

We can find f'(x).

⇒ f'(x) = d(x³ - 3x² - 24x + 5)/dx.

⇒ f'(x) = 3x² - 6x - 24.

⇒ f'(x) = 3(x² - 6x - 8).

⇒ f'(x) = 3(x² - 4x - 2x - 8).

⇒ f'(x) = 3(x(x - 4) - 2 ( x - 4)).

⇒ f'(x) = 3(x - 2)(x - 4).

Put f'(x) = 0, in equation, we get.

⇒ (x - 2)(x - 4) = 0.

⇒ x = 2  and x = 4.

To find f''(x) we get,

⇒ f''(x) = d(x² - 6x - 8)/dx.

⇒ f''(x) = 2x - 6.

Put the value of x in equation f''(x).

Put x = 2 in equation f''(x).

⇒ f''(x) = 2(2) - 6.

⇒ f''(x) = 4 - 6.

⇒ f''(x) = -2.

⇒ f''(x) < 0 at x = 2.

x is point of local maxima.

f(x) is maximum at x = 2.

Put x = 4 in equation f''(x), we get.

⇒ f''(x) = 2(4) - 6.

⇒ f''(x) = 8 - 6.

⇒ f''(x) = 2.

⇒ f''(x) > 0 at x = 4.

x is the point of local minima.

f(x) is minimum at x = 4.

Answer 2

$\large\underline{\red{\sf \orange{\bigstar} Correct\; Question}}$

• find the local maximum and local mimnimun value of f(x)= x ^{2} - 3x^2 - 24x + 5

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$\large\underline{\pink{\sf \green{\bigstar} Solution}}$⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Local maximum and local minimum of the value.

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• ::⇒ f(x) = x³ - 3x² - 24x + 5.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\pink{\sf{\star\;☯ \;As \;we\; know \;that,}}$

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$\red{\sf{\star\;☯ \;We \;can\; find\; f'(x).}}$

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• ::⇒ f'(x) = d(x³ - 3x² - 24x + 5)/dx.

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• ::⇒ f'(x) = 3x² - 6x - 24.

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• ::⇒ f'(x) = 3(x² - 6x - 8).

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• ::⇒ f'(x) = 3(x² - 4x - 2x - 8).

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• ::⇒ f'(x) = 3(x(x - 4) - 2 ( x - 4))

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• ::⇒ f'(x) = 3(x - 2)(x - 4)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\blue{\sf{\star\;⚛\; Put \;f'(x) \;=\; 0, \;in \;equation, \;we \;get}}$

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• ::⇒ (x - 2)(x - 4) = 0.

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• ::⇒ x = 2  and x = 4.

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$\pink{\sf{\star\;⚛ \;To\; find \;f''(x) \;we \;get ?}}$

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• ::⇒ f''(x) = d(x² - 6x - 8)/dx.

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• ::⇒ f''(x) = 2x - 6.${\boxed{\red{\checkmark{}}}}$

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Put the value of x in equation f''(x).

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$\purple{\sf{\star\;☯ \;Put\; x\; =\; 2 \;in \;equation\; f''(x)}}$

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• ::⇒ f''(x) = 2(2) - 6.

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• ::⇒ f''(x) = 4 - 6.

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• ::⇒ f''(x) = -2.

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• ::⇒ f''(x) < 0 at x = 2.${\boxed{\pink{\checkmark{}}}}$

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$\orange{\sf{\star\;☮ \;x \;is \;point \;of\; local\; maxima.}}$

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$\green{\sf{\star\;☮ \;f(x)\; is\; maximum \;at \;x \;=\; 2.}}$

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Put x = 4 in equation f''(x), we get.

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• ::⇒ f''(x) = 2(4) - 6.

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• ::⇒ f''(x) = 8 - 6.

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• ::⇒ f''(x) = 2.

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• ::⇒ f''(x) > 0 at x = 4.${\boxed{\blue{\checkmark{}}}}$

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$\large\underline{\red{\sf \purple{\bigstar} }}$x is the point of local minima.

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f(x) is minimum at x = 4.${\boxed{\green{\checkmark{}}}}$

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