What are the critical values of f(x)=x/sqrt(x^2+2)-(x-1)^2?

Nov 21, 2017

$x \cong 1.1629$

Explanation:

The critical values of a function can be found by looking at its derivative. If the derivative is either 0 or undefined, this is a critical value.

So, we first start by computing $f ' \left(x\right)$:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{x}{\sqrt{{x}^{2} + 2}} - {\left(x - 1\right)}^{2}\right)$
Let's split this into two derivatives:
$= \frac{d}{\mathrm{dx}} \left(\frac{x}{\sqrt{{x}^{2} + 2}}\right) - \frac{d}{\mathrm{dx}} \left({\left(x - 1\right)}^{2}\right)$
We start by computing the first expression:
$\frac{d}{\mathrm{dx}} \left(\frac{x}{\sqrt{{x}^{2} + 2}}\right)$
By the quotient rule, this is equal to
$= \frac{\frac{\mathrm{dx}}{\mathrm{dx}} \sqrt{{x}^{2} + 2} - x \frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} + 2}\right)}{\sqrt{{x}^{2} + 2}} ^ 2$

$= \frac{\sqrt{{x}^{2} + 2} - x \frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{2} + 2}\right)}{{x}^{2} + 2}$
We now have to solve for the derivative of $\sqrt{{x}^{2} + 2}$. If we let $u = {x}^{2} + 2$, then we can get to this expression using the chain rule:
$= \frac{d}{\mathrm{du}} \left(\sqrt{u}\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 2\right)$

$= \frac{2 x}{2 \sqrt{u}}$
Now let's substitute back in for $u$:
$= \frac{2 x}{2 \sqrt{{x}^{2} + 2}}$
So now we know that
$\frac{d}{\mathrm{dx}} \left(\frac{x}{\sqrt{{x}^{2} + 2}}\right) = \frac{\sqrt{{x}^{2} + 2} - x \left(\frac{2 x}{2 \sqrt{{x}^{2} + 2}}\right)}{{x}^{2} + 2}$

$= \frac{\sqrt{{x}^{2} + 2} - \frac{{x}^{2}}{\sqrt{{x}^{2} + 2}}}{{x}^{2} + 2} = \frac{\frac{\sqrt{{x}^{2} + 2} \sqrt{{x}^{2} + 2}}{\sqrt{{x}^{2} + 2}} - {x}^{2} / \left(\sqrt{{x}^{2} + 2}\right)}{{x}^{2} + 2}$

$= \frac{\frac{{x}^{2} + 2 - {x}^{2}}{\sqrt{{x}^{2} + 2}}}{{x}^{2} + 2} = \frac{2}{\left({x}^{2} + 2\right) \sqrt{{x}^{2} + 2}}$
And there we have the first part of the derivative. Now for the second:
$\frac{d}{\mathrm{dx}} \left({\left(x - 1\right)}^{2}\right)$
We can let $u = x - 1$
$= \frac{d}{\mathrm{du}} \left({u}^{2}\right) \frac{d}{\mathrm{dx}} \left(x - 1\right)$

$= 2 u$
Substituting back in, we get:
$= 2 \left(x - 1\right) = 2 x - 2$
Now we know the value of $f ' \left(x\right)$:
$f ' \left(x\right) = \frac{2}{\left({x}^{2} + 2\right) \sqrt{{x}^{2} + 2}} - 2 x + 2$

To find the critical points, we set this expression equal to 0 and solve for x.
$\frac{2}{\left({x}^{2} + 2\right) \sqrt{{x}^{2} + 2}} - 2 x + 2 = 0$

Now, this equation is a very hard one to solve, and I'm not the one to explain how to solve it. But I can tell you that the only real solution is around $1.1629$