# Verify by Second Derivative Test

$$A(x)=2\sqrt{x^2-16}+\frac14\sqrt{68x^2-x^4-256}\;,\;\; (4 < x < 8)$$

of which the derivative is:

$$a'(x)=\frac{2x}{\sqrt{x^2-16}}+\frac{136x-4x^3}{8\sqrt{68x^2-x^4-256}}$$

I first had to find a value of $x$ for which $A'(x)=0$ The result I got was a local maximum at $x=7.1296$ and a local minimum at $x=4$ The part I am struggling with is:

Verify by the second derivative test that this value of $x$ corresponds to a local maximum of $A(x)$ .

As I understand I have to find the derivative of the above function, then find the derivative of that and then input $x=7.1296$ to solve the equation and if the answer is $<0$ then the local maximum is at $x=7.1296$ . Please let me know if I have the correct method. If so, then I'm struggling to get the second derivative and need help with that.

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The lack of necessary parenthesis makes this incomprehensible. Are you taking the square root of the whole of $x^2 - 16$, $68x^2 - x^4 - 256$ or something else? –  Lee Yiyuan Feb 25 '14 at 14:23