 # Find the extreme values (absolute and local) of the function Irvin Dukes 2021-12-10 Answered
Find the extreme values (absolute and local) of the function over its natural domain, and where they occur.
$y=x-4\sqrt{x}$
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Given,
$y=x-4\sqrt{x}$
The domain of given function is $\left[0,\mathrm{\infty }\right)$.
Differentiating with respect to x, we get
$\frac{dy}{dx}=1-4\cdot \frac{1}{2\sqrt{x}}$
$=1-\frac{2}{\sqrt{x}}$
Now critical points are the points at which first derivative is zero or not defined.
So to find critical points, solve the first derivative by equating it to zero. That is,
$\frac{dy}{dx}=0$
$⇒1-\frac{2}{\sqrt{x}}=0$
$⇒\frac{2}{\sqrt{x}}=1$
$⇒\sqrt{x}=2$
$⇒x=4$
& also the first derivative is not defined at x=0.
Therefore critical values are x=0 & x=4.
Step 2
Now,
$\frac{dy}{dx}<0$ to the left of x=4 & $\frac{dy}{dx}>0$ to the right of x=4.
Therefore x=4 is the point of local minima & local minimum value is,
$y=4-4\sqrt{4}$
=4-8
=-4
Therefore local minimum and absolute minimum value is -4 occurs at x=4.