Question

Find maximum/minimum value of $y$ in $y=x^3-6x^2+9x+15$

• A. $y_{min}=7.5$ at $x=3$ and $y_{max}=9.5$ at $x=1$
• B. $y_{min}=15$ at $x=3$ and $y_{max}=19$ at $x=1$
• C. $y_{min}=30$ at $x=3$ and $y_{max}=38$ at $x=1$
• D. $y_{min}=45$ at $x=3$ and $y_{max}=57$ at $x=1$

Answer 1

The correct option is B $y_{min}=15$ at $x=3$ and $y_{max}=19$ at $x=1$

Given, $y=x^3-6x^2+9x+15$

$\therefore y^{\prime }=3x^2-12x+93x^2-12x+9=03x^2-3x-9x+9=03x(x-1)-9(x-1)=0(3x-9)(x-1)=0x=3,x=1$

Now $y^{\prime \prime }=6x-12$

$y^{\prime \prime }\left(3\right)>0$ Hence minimum value will be found at $x=3$ and

$y^{\prime \prime }\left(1\right)<0$ Hence maximum value will be found at $x=1$
By putting the values of $x$ we get

$y_{min}=15$ at $x=3$ and $y_{max}=19$ at $x=1$