Find the maximum and minimum value of following function. 2 x^3-1… - Vidyadip

Question

Find the maximum and minimum value of following function. $2 x^3-15 x^2+36 x+10$

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Answer

Suppose $ y=2 x^3-15 x^2+36 x+10 $
So,$\frac{d y}{d x}=6 x^2-30 x+36$
and $ \frac{d^2 y}{d x^2}=12 x-30$
For maxima and minima $\frac{d y}{d x}=0$
$\text { So, } 6 x^2-30 x+36=0$
$\Rightarrow 6\left(x^2-5 x+6\right)=0$
$\Rightarrow x^2-5 x+6=0$
$\text { or } \quad(x-3)(x-2)=0$
$\Rightarrow x=2,3$
now at $ x=2, \frac{d^2 y}{d x^2}=12 \times 2-30$
$=-6 < 0$
Function will maximum at $x=2$ and maximum value of function
$=2(2)^3-15(2)^2+36(2)+10$
$=16-60+72+10$
$=38$
again at $x=3, \frac{d^2 y}{d x^2}=12 \times 3-30$
$=6>0$
Function will be minimum at $x=3$ and minimum value of function
$=2(3)^3-15(3)^2+36(3)+10$
$=2 \times 27-15 \times 9+108+10$
$=54-135+108+10=37$

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