Answer:
[tex]\displaystyle \text{min} = \frac{\sqrt{3+2\sqrt{3}}}{2} - \frac{2\sqrt[4]{3}}{\sqrt{2}} \text{ at } x = \frac{\sqrt{3}}{2}\text{ and } \\ \\ \text{max} = 2\sqrt{5} -4 \text{ at } x = 4[/tex]
Step-by-step explanation:
We want to find the maximum and minimum values of the function:
[tex]\displaystyle f(x) = \sqrt{x + x^2} - 2\sqrt{x}[/tex]
On the interval [0, 4].
First, evaluate its endpoints:
[tex]\displaystyle \begin{aligned} f(0) &= \sqrt{(0)+(0)^2} - 2\sqrt{0} \\ &= 0 \\ \\ f(4) &= \sqrt{(4)+(4)^2} - 2\sqrt{(4)} \\ &= 2\sqrt{5} -4 \end{aligned}[/tex]
Recall that the extrema of a function occurs at its critical points; that is, where its derivative equals zero (or is undefined).
Take the derivative of both sides:
[tex]\displaystyle f'(x) = \frac{d}{dx}\left[ \sqrt{x + x^2} - 2\sqrt{x}\right][/tex]
Differentiate:
[tex]\displaystyle \begin{aligned} f'(x) &= \frac{1}{2\sqrt{x + x^2}} \cdot (1 + 2x) - 2\left(\frac{1}{2\sqrt{x}}\right) \\ \\ &= \frac{2x+1}{2\sqrt{x+x^2}} - \frac{1}{\sqrt{x}} \\ \\\end{aligned}[/tex]
Note that the derivative is undefined at x = 0. Hence, x = 0 is a critical point.
Solve for the zeros of the derivative:
[tex]\displaystyle\begin{aligned} \frac{2x+1}{2\sqrt{x + x^2}} - \frac{1}{\sqrt{x}} &= 0\\ \\ \frac{2x+1}{2\sqrt{x}\sqrt{1 + x }} - \frac{1}{\sqrt{x}} &= 0 \\ \\ \frac{2x+1}{2\sqrt{1+x}} - 1 &= 0\\ \\ 2x + 1 &= 2\sqrt{1+x} \\ \\ 4x^2 + 4x + 1 &= 4 + 4x \\ \\ x^2 &= \frac{3}{4} \\ \\ x= \frac{\sqrt{3}}{2} \end{aligned}[/tex]
Therefore, our only two critical points are at x = 0 and x = √3/2:
Evaluate the function at x = √3/2:
[tex]\displaystyle \begin{aligned} f\left(\frac{\sqrt{3}}{2}\right) &= \sqrt{\left(\frac{\sqrt{3}}{2} \right)+ \left(\frac{\sqrt{3}}{2}\right)^2} - 2 \sqrt{\left(\frac{\sqrt{3}}{2}\right)} \\ \\ &= \frac{\sqrt{3+2\sqrt{3}}}{2}- \frac{2\sqrt[4]{3}}{\sqrt{2}} \\ \\ &\approx -0.5900\end{aligned}[/tex]
In conclusion: the exact maximum and minimum values of f on the interval [0, 4] is:
[tex]\displaystyle \text{min} = \frac{\sqrt{3+2\sqrt{3}}}{2} - \frac{2\sqrt[4]{3}}{\sqrt{2}} \text{ at } x = \frac{\sqrt{3}}{2}\text{ and } \\ \\ \text{max} = 2\sqrt{5} -4 \text{ at } x = 4[/tex]
