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Find the absolute maximum and minimum values of a function f given by f(x) = 2x3 − 15x2 + 36x + 1 on the interval [1, 5]. - Mathematics

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प्रश्न

Find the absolute maximum and minimum values of a function f given by f(x) = 2x3 − 15x2 + 36x + 1 on the interval [1, 5].

Find the absolute maximum and absolute minimum of function f(x) = 2x3 − 15x2 + 36x + 1 on [1, 5].

योग
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उत्तर

Given: f(x) = 2x3 − 15x2 + 36x + 1

Differentiate w. r. t. x,

f'(x) = 6x2 − 30x + 36

Put f'(x) = 0

6(x2 − 5x + 6) = 0

x2 − 5x + 6 = 0

x2 − 3x − 2x + 6 = 0

x(x − 3) −2(x − 3) = 0

(x − 3)(x − 2) = 0

then, x =3 or 2

x = 1, 2, 3, 5

Now f(1) = 2(1)3 – 15(1)2 + 36(1) + 1

= 2 – 15 + 36 + 1

= 24

f(2) = 2(2)3 – 15(2)2 + 36(2) + 1

= 16 – 60 + 72 + 1

= 29

f(3) = 2(3)3 – 15(3)2 + 36(3) + 1

= 54 – 135 + 108 + 1

= 28

f(5) = 2(5)3 – 15(5)2 + 36(5) + 1

= 250 – 375 + 180 + 1

= 56

Hence, the absolute maximum value when x = 5 is 56, and the absolute minimum value when x = 1 is 24.

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अध्याय 18: Maxima and Minima - Exercise 18.4 [पृष्ठ ३७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 18 Maxima and Minima
Exercise 18.4 | Q 5 | पृष्ठ ३७

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