The minimum value

Question:

The minimum value of $\mathrm{f}(\mathrm{x})=x 4-x 2-2 x+6$ is

(a) 6

(b) 4

(c) 8

(d) none of these

Solution:

(b) 4

Given : $f(x)=x^{4}-x^{2}-2 x+6$

$\Rightarrow f^{\prime}(x)=4 x^{3}-2 x-2$

$\Rightarrow f^{\prime}(x)=(x-1)\left(4 x^{2}+4 x+2\right)$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow(x-1)\left(4 x^{2}+4 x+2\right)=0$

$\Rightarrow(x-1)=0$

$\Rightarrow x=1$

Now,

$f^{\prime \prime}(x)=12 x^{2}-2$

$\Rightarrow f^{\prime \prime}(1)=12-2=10>0$

So, $x=1$ is a local minima.

The local minimum value is given by

$f(1)=1-1-2+6=4$

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