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

1 Answer
Nov 21, 2017

Answer:

#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#