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

of which the derivative is:


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.

  • $\begingroup$ 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? $\endgroup$ – Yiyuan Lee Feb 25 '14 at 14:23
  • $\begingroup$ You can use dollar symbols ("$") to input mathematical formulas. See this: math.stackexchange.com/editing-help#latex $\endgroup$ – frabala Feb 25 '14 at 14:31
  • $\begingroup$ My apolgies I was having trouble entering it. I have edited the function and it is now correct. $\endgroup$ – skeeto Feb 25 '14 at 15:29
  • $\begingroup$ I am not sure whether this helps, but maybe some parts of this expression could be simplified a little using $x^4-65x^2+256=(x^2-4)(x^2-64)$. $\endgroup$ – Martin Sleziak Feb 25 '14 at 15:34

Yes, your sketched approach is just fine. I'll write your first derivative using the exponent $1/2$ to replace the squareroot symbol, and I also factored out $4$ from the numerator to simplify.

$$A'(x) = \dfrac{2x}{(x^2-16)^{1/2}} + \frac {34x - x^3}{2(68x^2 - x^4 - 256)^{1/2}}$$

You might want to try combining the two terms by finding a common denominator. Then to find $A''(x)$, you'll only need to use the quotient rule once.

$$A'(x)= \frac{4x(68x^2-x^4-256)^{1/2}+(34x-x^3)(x^2 - 16)^{1/2}}{2\Big((x^2-16)(68x^2 - x^4 -256)\Big)^{1/2}}$$

It's hard to say whether using the quotient rule twice will be easier than using the quotient rule once but needing the product rule a number of times. Either way, it may seem very messy, but you need to simply persevere, being careful along the way. (Personally, I'd go with the "original" $A'(x)$.

Hang in there, and feel free to check back if you need any verification.

  • $\begingroup$ The function you've taken is incorrect which was probably an error on my part in the way I typed it. I have edited the function and it appears as it should now $\endgroup$ – skeeto Feb 25 '14 at 15:30
  • $\begingroup$ Note: I simply factored out $4$ from the numerator of the second term in your sum: $$136x-4x^3 = 4(34 x - x^3)$$ $$A'(x)=\frac{2x}{\sqrt{x^2-16}}+\frac{136x-4x^3}{8\sqrt{68x^2-x^4-256}} = \frac{2x}{\sqrt{x^2 - 16}} + \frac{\color{blue}{4}(34x - x^3)}{\color{blue}{8}\sqrt{68x^2 - x^4 - 256}}$$Then, $\frac 48$ reduces to $\frac 12$. $\endgroup$ – amWhy Feb 25 '14 at 15:34
  • $\begingroup$ Also note that $\sqrt{f(x)} = (f(x))^{1/2}$. So "my" A'(x) is precisely the same A'(x) you posted (before, and now). I made out your earlier notation just fine. ;-) $\endgroup$ – amWhy Feb 25 '14 at 15:40

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