Question

Let $y=(x+\frac{4}{x^2})$ and $x\in +R$

The minimum value of $y$ is

• A. $-4$
• B. $3$
• C. $8$
• D. $7$

Answer 1

The correct option is B $3$

Global maximum is the maximum value of the function over the entire domain and not just between a region.

The given information is:

$y=x+\frac{4}{x^2}$

Differentiating the function w.r.t.$x$ and equating it to 0 to find the extremum points.

$\Rightarrow y^{\prime }=1-\frac{8}{x^3}$

Now $y^{\prime }=0$

$\Rightarrow 1-\frac{8}{x^3}=0$

$\Rightarrow x^3=8$

$\Rightarrow x=2$

To find whether the function has the maximum or minimum value at the given extremum point we find the double derivative of the function. If its is greater than zero at the extremum point then its a minimum point else a maximum point. The double derivative test fails if it is equal to zero.

So find the double derivative by differentiating $y^{\prime }$ we get,

$y^{\prime \prime }=\frac{24}{x^4}$

$\Rightarrow y^{\prime \prime }\left(2\right)=\frac{24}{16}$

$\Rightarrow y^{\prime \prime }>0$. Hence the point $x=2$ is a point of minima.

The minimum value of the function is

$\Rightarrow y\left(2\right)=2+\frac{4}{4}$

$\Rightarrow y\left(2\right)=3$ ......Answer

Thus the minimum value of the function is 3.