Question

Find absolute maxima and absolute minima of f(x)=2x^3-15x^2+36x+10 on [1,4] . ​

Answer 1

Answer:

$= > y = f(x) = 2 {x}^{3} - 15 {x}^{2} + 36x + 10 \\ = > \frac{dy}{dx} = 6 {x}^{2} - 30x + 36 \\ \\ let \: \frac{dy}{dx} = 0 \\ = > 6 {x}^{2} - 30x + 36 = 0 \\ = > 6( {x}^{2} - 5x + 6) = 0 \\ = > {x}^{2} - 5x + 6 = 0 \\ = > {x}^{2} - 2x - 3x + 6 = 0 \\ = > x(x - 2) - 3(x - 2) = 0 \\ = > (x - 2)(x - 3) = 0 \\ = > x = 2 \: or \: 3 \\ \\ \\ f(1) = 2 {(1)}^{3} - 15 {(1)}^{2} + 36(1) + 10 = 2 - 15 + 36 + 10 = 48 - 15 = 33 \\ f(2) = 2 {(2)}^{3} - 15 {(2)}^{2} + 36(2) + 10 = 16 - 60 + 36 + 10 = 62 - 60 = 2 \\ f(3) = 2 {(3)}^{3} - 15 {(3)}^{2} + 36(3) + 10 = 54 - 135 + 108 + 10 = 172- 135 = 37 \\ f(4) = 2 {(4)}^{3} - 15 {(4)}^{2} + 36(4) + 10 = 128 - 240 + 144 + 10 = 282 - 240 = 42 \\ \\ \\ \\ \\ \\ hence \: \: \: absolute \: \: maxima \: \: = 42 \: \: and \: \: absolute \: \: minima \: \: = 2$

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