Question

The function $y=x^4-6x^2+8x+11$ has a minimum at $x=$

• A. 1
• B. -2
• C. 3
• D. 4

Answer 1

Answer: (B)

We have, $y=x^4-6x^2+8x+11$
For maximum or minimum $\frac{dy}{dx}=0$
$\therefore \frac{dy}{dx}=4x^3-12x+8=0$ i.e., $x^3-3x+2=0$
$\Rightarrow \left(x-1\right)\left(x^2+x-2\right)=0\Rightarrow x=1,-2$
Now, $\frac{d^{2}y}{d{x}^2}=12x^2-12=12\left(x^2-1\right)$
$\frac{d^{2}y}{d{x}^2}{|}_{atx=0}=0$
Neither minimum nor maximum point
and $\frac{d^{2}y}{d{x}^2}{|}_{atx=-2}=36>0$
$\therefore$ The given function has a minimum at
$x=-2$