+170 Let \(a\) and \(b\) be nonnegative real numbers such that \(a^2 + b^2 = 1\) Find the set of all possible values of \(a+b\)
+6248 \(\text{We can let $a=\cos(x),~b=\sin(x)$}\\ r (x)= \cos(x)+\sin(x)\\ \dfrac{dr}{dx} = -\sin(x)+\cos(x)\\ \dfrac{dr}{dx} =0 \Rightarrow x = \dfrac{\pi}{4}, ~\dfrac{5\pi}{4}\\ \dfrac{d^2r}{dx^2} = -\cos(x)-\sin(x)\\ \left. \dfrac{d^2r}{dx^2}\right|_{x=\frac \pi 4} <0 \text{ so this is the maxima}\\ \left. \dfrac{d^2r}{dx^2}\right|_{x=\frac {5\pi} 4} >0 \text{ so this is the minima}\\ r\left(\frac \pi 4\right) = \sqrt{2}\\ r\left(\frac{5\pi}{4}\right) = -\sqrt{2}\\ \text{and we get that $a+b \in [-\sqrt{2},\sqrt{2}]$} \)
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