Suppose we have $$f(x)=\sqrt{x-1}+\sqrt{5-x}$$; how do we find range for this function? Single radicals are easy , but two of them are in this particular function.
I have the domain of the function and I can only think of differentiation to get the maximum of the function in the valid domain to find the range of $$f(x)$$. Is differentiation the only way, or is there something easier ?

• Maximum is when $x=3$ Mar 13, 2019 at 19:54
• Its evident from the graph Mar 13, 2019 at 20:07
• Differentiation will easily lead to the conclusion that the largest number in the range occurs at $x=3$, so why not use it? Mar 13, 2019 at 20:11
• found the answer that way , just trying to find other methods , since I am new to calculus. Mar 13, 2019 at 20:14

By the AM-GM inequality, $$\sqrt{2+t} \sqrt{2-t} \le \frac {(2+t)+(2-t)} 2 = 2,$$
so $$(\sqrt{2+t}+\sqrt{2-t})^2 = (2+t)+(2-t)+2\sqrt{2+t}\sqrt{2-t}=4+2\sqrt{2+t}\sqrt{2-t}\le 8$$
so $$\sqrt{2+t}+\sqrt{2-t}\le\sqrt{8}.$$
Now let $$x=t+3.$$
• One way to possibly motivate this (I was curious, as seemed pulled out of a hat) is to manipulate the original equation into $(y^2 - 4)^2 = 4(x-1)(5-x)$ $(y$ is the dependent variable) where $1 \leq x \leq 5.$ Since the max and min of $y$ are determined by those of $(y^2 - 4)^2,$ it suffices to locate the max and min of the product $(x-1)(5-x),$ which can be viewed as the area of a rectangle with dimensions $3+u$ and $3-u$ for $0 \leq u \leq 2,$ and it's well known that areas of rectangles (of fixed perimeters) are maximized when squares and minimized when "furthest from a square" (via AM-GM). Mar 14, 2019 at 7:58