Math, asked by Aribaaa34, 11 months ago

अन्तराल [0, 3] पर 3x^{4} – 8x^{3} + 12x^{2} – 48x + 25 के उच्चतम मान और निम्नतम मान ज्ञात कीजिए।

Answers

Answered by Anonymous
1

Answer:

f (x) = 3x4 – 8x3 + 12x2 – 48 x + 25

f’(x) = 12x3 – 24x2 + 24x – 48

f‘(x) = 0 ⇒ 12 (x3 – 2x2 + 2x - 4) = 0

⇒ 12 (x2 (x – 2) + 2 (x – 2))

⇒ 12 ( (x – 2) (x2 + 2) ) = 0 (x – 2)(x2 + 2) = 0

⇒ x = 2 but x2 + 2 ≠ 0

The points are f(0), f(2), f(3)

f(x) = 3x4 – 8x3 + 12x2 – 48x + 25

f(0) = 25

f(2) = 3(16) – 8(8) + 12(4) – 48(2) + 25 = 48 – 64 + 48 – 96 + 25 = -39

f(3) = 3(34) – 8 (33) + 12(32) – 48(3) + 25 = 243 – 216 + 108 – 144 + 25

376 – 360 = 16

∴ maximum of f (x) at x = 0 is 25

minimum of f (x) at x = 2 is – 39.

Answered by amitnrw
0

Given :  f(x) = 3x⁴ - 8x³  + 12x²  - 48x  + 25

To find :  अन्तराल [0, 3] पर उच्चतम मान और निम्नतम मान ज्ञात कीजिए

Solution:

f(x) = 3x⁴ - 8x³  + 12x²  - 48x  + 25

f'(x) =  12x³ - 24x²  + 24x - 48

= 12(x³  - 2x²  + 2x - 4)

= 12 (x  - 2) ( x² + 2)

f'(x) = 12 (x  - 2) ( x² + 2)

 f'(x) = 0

=> 12 (x  - 2) ( x² + 2) = 0

=> x = 2   x² + 2  ≠ 0

f'(x) = 12x³ - 24x²  + 24x - 48

f''(x) = 36x² - 48x  + 24

f''(x) = 12(3x²  - 4x + 2)

x = 2

f''(x) = 72 > 0

=> x = 2  निम्नतम मान

f(2) = 3*2⁴ - 8*2³  + 12*2²  - 48*2  + 25

= 48 - 64 + 48 - 96+ 25

= -39

निम्नतम मान = -39

अन्तराल [0, 3]

f(0) =  3*0⁴ - 8*0³  + 12*0²  - 48*0  + 25 = 25

f(3) =  3*3⁴ - 8*3³  + 12*3²  - 48*3  + 25  

= 243 - 216 + 108 - 144 + 25

= 16

25 > 16

 उच्चतम मान = 25

उच्चतम मान = 25   निम्नतम मान = -39   अन्तराल [0, 3] पर  f(x) = 3x⁴ - 8x³  + 12x²  - 48x  + 25

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