Find the maximum and minimum values of function $y = 2 x^{3} - 15 x^{2} + 36 x + 11$

42; 41

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39; 38

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35; 34

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52; 51

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Solution

The correct option is B $39; 38$
$y = 2 x^{3} - 15 x^{2} + 36 x + 11$
$\frac{d y}{d x} = 6 x^{2} - 30 x + 36$
$\frac{d y}{d x} = 0 = 6 x^{2} - 30 x + 36 = 0$
$x^{2} - 5 x + 6 = 0$
$x^{2} - 2 x - 3 x + 6 = 0$
$(x - 3) (x - 2) = 0$
Roots of the quadratic equation are
$x = 2, 3$

Now double differentiation of given equation
$\frac{d^{2} y}{d x^{2}} = 12 x - 30$
At $x = 2$ , $\frac{d^{2} y}{d x^{2}} = - 6$
Hence at $x = 2$ there is maxima of value $y_{max} = 39$
At $x = 3$ , $\frac{d^{2} y}{d x^{2}} = + 6$
Hence at $x = 3$ , there is minima of value $y_{min} = 38$

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