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, $\frac{d^{2}y}{dx^{2}} = 12x^{2} - 12 = 12\left(x^{2} - 1\right)$
$\frac{d^{2} y}{dx^{2}}|_{at\, x=0} =0$
Neither minimum nor maximum point
and $\frac{d^{2}y}{dx^{2}}| _{at\, x=-2} =36 > 0$
$\therefore$ The given function has a minimum at
$x = - 2$