# Chapter 4 - Applications of the Derivative - 4.2 Extreme Values - Exercises - Page 181: 41

The maximum is $f(4) \approx0.47213$ and the minimum is $f(\sqrt{3}/2 ) \approx -0.589980$

#### Work Step by Step

Given $$y=\sqrt{x+x^{2}}-2 \sqrt{x}, \quad[0,4]$$ Since \begin{align*} \frac{dy}{dx}&=\frac{1+2x}{2\sqrt{x+x^2}}-\frac{1}{\sqrt{x}}\\ &=\frac{1+2 x-2 \sqrt{1+x}}{2 \sqrt{x} \sqrt{1+x}} \end{align*} Find the critical points \begin{align*} f'(x)&=0\\ \frac{1+2 x-2 \sqrt{1+x}}{2 \sqrt{x} \sqrt{1+x}}&=0\\ 1+2 x-2 \sqrt{1+x}&=0\\ 4x^2-3&=0 \end{align*} Then $x=\pm\sqrt{3}/2$. Because $-\sqrt{3}/2\notin [0,4]$, then $f(x)$ has a critical point $\sqrt{3}/2$, since \begin{align*} f(0)&=0\\ f(\sqrt{3}/2 )&\approx -0.589980\\ f(4)&\approx0.47213 \end{align*} Then the maximum is $f(4) \approx0.47213$ and the minimum is $f(\sqrt{3}/2 ) \approx -0.589980$

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