Min value of $F(x) = x - \sqrt{1-x^2} $

Question

Min value of $F(x) = x - \sqrt{1-x^2} $

But using derivatives $F(x) = x - \sqrt{1-x^2} $

i stucked at $2\sqrt{1-x^2} + 2x = 0$

$\sqrt{1-x^2} > 0 $

I get $-1<x<1$

Please help me? is there another way to solve it btw?

Answer 1

I think there may be something wrong in this (that means, your derivative): $2\sqrt{1-x^2} + 2x = 0$

My answer:

First let's notice that the function domain is $[-1,1]$ for that $1-x^2\geqslant 0$. Then it is easy to see $f'(x)=0$ when $f(x)$ reaches to minimum. So that $$ f'(x)=\frac{x}{\sqrt{1-x^2}}+1 $$ set $f'(x)=0$ we get $x=-\frac{1}{\sqrt{2}}$, and $f(x)=-\sqrt{2}$ where it is the minimum. Rigorously speaking, you need to check the derivative on both side to judge whether it is the min value, but in this question it is obvious.

Answer 2

Okay so if $F(x) = x - \sqrt{1-x^2} = x - (1-x^2)^{\frac 12}$ then $F'(x) =1 - \frac 12(1-x^2)^{-\frac 12}*(-2x) = 1+\frac {x}{\sqrt{1-x^2}}$

Solving for $1+\frac {x}{\sqrt{1-x^2}}=0$ yield $x =-\sqrt{1-x^2}$ so $x^2 = 1-x^2; x \le 0$. Or $2x^2 =1$ or $x=-\frac {\sqrt 2}2$ and

$F(-\frac {\sqrt 2}2) = -\frac {\sqrt 2}2-\sqrt{1-\frac 12}= -\frac {\sqrt 2}2 - \frac {\sqrt 2}2 = -\sqrt 2$.

Yegawds. I made several arithmetic errors getting here!

Answer 3

A function $f(x)$ can have its extrema at $x$ values that satisfy $f'(x)=0$. Thus:

\begin{align} F(x)&=x-\sqrt{1-x^2}\\ F'(x)&=1+\frac{x}{\sqrt{1-x^2}}\\ 0&=\frac{\sqrt{1-x^2}+x}{\sqrt{1-x^2}}\\ 0&=\sqrt{1-x^2}+x\quad\textrm{(notice here that $x$ has to be negative)}\\ -x&=\sqrt{1-x^2}\\ x^2&=1-x^2\\ 2x^2&=1\\ x^2&=\frac12\\ x&=-\frac1{\sqrt2} \end{align}

Since the domain is $x\in[-1,1]$, we can test $F(-1)$, $F(1)$, and $F(-\frac1{\sqrt2})$. You should be able to figure that $F(-\frac1{\sqrt2})$ will give you the minimum value.

Answer 4

If the usage of calculus is not mandatory,

as $1-x^2\ge0,$ WLOG $x=\sin t,$ where $-\dfrac\pi2\le t\le\dfrac\pi2$

$$f(x)=\sin t-\cos t=\sqrt2\sin\left(t-\dfrac\pi4\right)$$

Now $-\dfrac\pi2-\dfrac\pi4\le t-\dfrac\pi4\le\dfrac\pi2-\dfrac\pi4$

$\implies-1\le\sin\left(t-\dfrac\pi4\right)\le\dfrac1{\sqrt2}$