# Math Help - Finals question help! (calculus)

1. ## Finals question help! (calculus)

Hi everyone,

My Calculus teacher gave us some review sheets of what to expect on the final. I've narrowed it down to this question that I can't figure out no matter what. Can anyone help me? Here is the problem:

Find the value of x where f(x)= x^2-3(x+2)^(1/2) has its absolute minimum.

I know I'm supposed to take the derivative and set that equal to zero to find the critical points. However, I keep getting stuck and can't seem to find the critical points because I get lost in the math.

Thank you for any help!

2. Originally Posted by rIBBON:lacedx
Hi everyone,

My Calculus teacher gave us some review sheets of what to expect on the final. I've narrowed it down to this question that I can't figure out no matter what. Can anyone help me? Here is the problem:

Find the value of x where f(x)= x^2-3(x+2)^(1/2) has its absolute minimum.

I know I'm supposed to take the derivative and set that equal to zero to find the critical points. However, I keep getting stuck and can't seem to find the critical points because I get lost in the math.

Thank you for any help!
I will assume that by absolute minimum you mean global minimum.

$f(x)=x^2-3(x+2)^{1/2}$

$f'(x)=2x-3(1/2)(x+2)^{-1/2}=2x-\frac{3}{2(x+2)^{1/2}}$

setting $f'(x)=0$ gives us:

$2x-\frac{3}{2(x+2)^{1/2}}=0\ \ \ \ \ ..(1)$

simplifying:

$\frac{4}{3}x=\frac{1}{(x+2)^{1/2}}$

or after squaring:

$\frac{16}{9}x^2 (x+2)-1=0$

Now this is a cubic, which has three roots, of these two are spurious (that
is are not zeros of the derivative of $f(x)$, and the other is
a genuine root. It is probably better to go for a direct numerical solution of
$(1)$ rather than solve this cubic.

A quick sketch of $f(x)$ shows the minimum is close to $x=0.5$,
and formula itteration of:

$x_{n+1}=\frac{3}{4(x_n+2)^{1/2}}$

gives the minimum occurs at $x \approx 0.4766$ and is $\approx -4.494$

RonL