Optimization problem PLEASE HELP?!?!?!

what is the critical number of f(x)=sqrt((x^(2)+36))+ sqrt((x-10)^2+144)
please for the love of god help me!!!!!! whats the answer?????

f(x)=sqrt(x^2+36) +{(x-10)^2+144}

Domain of f(x) is the entire set of real numbers because the expressions under the roots contain sum of a squared term and a positive number which cannot go negative for any real number.

Differentiating f with respect to x,
f'(x)=d/dx [{x^2+36}^(1/2) +{(x-10)^2+144}^(1/2)]

=(1/2)[{x^2+36}^((1/2)-1)] [d/dx {x^2+36}] +(1/2)[{(x-10)^2+144}^((1/2)-1)] [d/dx{(x-10)^2+144}]

=(1/2)[{x^2+36}^(-1/2)](2x) +(1/2)[{(x-10)^2+144}^(-1/2)] {2(x-10)}

=x/[sqrt{x^2+36}] +(x-10)/[sqrt{(x-10)^2+144}]

Critical numbers are those values of x in the domain of f(x) for which either f'(x) does not exist or for which f'(x)=0.

Here, f'(x) exists for all x in the domain of f(x).

Now setting f'(x)=0, we have
x/[sqrt{x^2+36}] =-(x-10)/[sqrt{(x-10)^2+144}]

Squaring on both sides,
x^2/[x^2+36] =(x-10)^2/[(x-10)^2+144]

Cross multiplying,
x^2[(x-10)^2+144] =(x-10)^2[x^2+36]

x^2(x-10)^2 +144x^2
=x^2(x-10)^2 +36(x-10)^2

Subtracting x^2(x-10)^2 from both sides,
144x^2 =36(x-10)^2

Dividing both sides by 36,
4x^2=(x-10)^2
i.e., 4x^2 =x^2-20x+100
i.e., 3x^2 +20x-100=0

Solving for x, we get
x=10/3 or x=-10.

Thus, the required critical numbers are 10/3 and -10.

There is a minimum at x = 10/3.