Question

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

Answer 1

`x~=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'(x)`:

`f'(x)=d/dx(x/(sqrt(x^2+2))-(x-1)^2)`
Let's split this into two derivatives:
`=d/dx(x/sqrt(x^2+2))-d/dx((x-1)^2)`
We start by computing the first expression:
`d/dx(x/sqrt(x^2+2))`
By the quotient rule, this is equal to
`=(dx/dxsqrt(x^2+2)-xd/dx(sqrt(x^2+2)))/(sqrt(x^2+2))^2`

`=(sqrt(x^2+2)-xd/dx(sqrt(x^2+2)))/(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:
`=d/(du)(sqrt(u))d/dx(x^2+2)`

`=(2x)/(2sqrt(u))`
Now let's substitute back in for `u`:
`=(2x)/(2sqrt(x^2+2))`
So now we know that
`d/dx(x/sqrt(x^2+2))=(sqrt(x^2+2)-x((2x)/(2sqrt(x^2+2))))/(x^2+2)`

`=(sqrt(x^2+2)-(x^2)/(sqrt(x^2+2)))/(x^2+2)=((sqrt(x^2+2)sqrt(x^2+2))/(sqrt(x^2+2))-x^2/(sqrt(x^2+2)))/(x^2+2)`

`=((x^2+2-x^2)/(sqrt(x^2+2)))/(x^2+2)=2/((x^2+2)sqrt(x^2+2))`
And there we have the first part of the derivative. Now for the second:
`d/dx((x-1)^2)`
We can let `u=x-1`
`=d/(du)(u^2)d/dx(x-1)`

`=2u`
Substituting back in, we get:
`=2(x-1)=2x-2`
Now we know the value of `f'(x)`:
`f'(x)=2/((x^2+2)sqrt(x^2+2))-2x+2`

To find the critical points, we set this expression equal to 0 and solve for x.
`2/((x^2+2)sqrt(x^2+2))-2x+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`