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प्रश्न
Examine the maxima and minima of the function f(x) = 2x3 - 21x2 + 36x - 20 . Also, find the maximum and minimum values of f(x).
Find the maximum and minimum value of the function:
f(x) = 2x3 - 21x2 + 36x - 20
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उत्तर १
`f(x)=2x^3-21x^2+36x-20`
`f^'(x)=6x^2-42x+36`
For finding critical points, we take f'(x)=0
`therefore 6x^2-42x+36=0`
`x^2-7x+6=0`
(x-6)(x-1)=0
For finding the maxima and minima, find f''(x)
f'(x)=12x-42
for x=6
f''(6)=30>0
Minima
for x=1
f''(1)=-30<0
Maxima
Maximum values of f(x) for x=1
f(1)=-3
minimum values of f(x) for x=6
f(6)=-128
∴ the maximum values of the function is -3 and the minimum value of the function is -128.
उत्तर २
f(x) = 2x3 - 21x2 + 36x - 20
∴ f'(x) = `2(3x^2) - 21(2x) + 36(1) - 1`
`= 6x^2 - 42x + 36 = 6(x^2 - 7x + 6)`
= 6(x - 1)(x - 6)
f has a maxima/minima if f'(x) = 0
i.e if 6(x - 1)(x-6) = 0
i.e if x - 1= 0 or x - 6 = 0
i.e if x = 0 or x = 6
Now f"(x) = 6(2x) - 42(1) = 12x - 42
∴ f"(1) = 12(1) - 42 = -30
∴ f"(1) < 0
Hence, f has a maximum at x = 1, by the second derivative test.
Also f"(6) = 12(6)- 42 = 30
∴ f"(6) > 0
Hence, f has a minimum at x = 6, by the second derivative test.
Now, the maximum value of f at 1,
`f(1) = 2(1^3) - 21(1^2) + 36(1) - 20`
= 2-21+36-20 = -3
and minimum value of f at x = 6
`f(6) = 2(6^3) - 21(6^2) + 36(1) - 20`
= 432 - 756 + 216 - 20 = -128
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