Can anyone help me find an inflection point for the following function without using graphing calculator.

Determine the intervals of concavity and points of inflection for the function


for my first derivative I did


for my common denominator I got


for my second derivative I got


But I am having trobule finding the inflection point.

  • $\begingroup$ The first step to finding an inflection point is locating where the second derivative equals zero. For what values of $x$ does $f''(x)=0$? $\endgroup$ – Ian Coley Mar 17 '13 at 18:13

First, simplify your $f'(x)$ before taking the second derivative.

$$f'(x)= \dfrac{(\sqrt{25-x^2})^2 - (x^2)}{\sqrt{25-x^2}} $$ $$ = \dfrac{25 - 2x^2}{\sqrt{25-x^2}}$$

Now use the quotient rule to compute $f''(x)$.

(You need to correct your $f''(x)$ and simplify, to find the common denominator), then determine when $f''(x) = 0$. That's where an inflection point would be.

Of course, we also need to check $x = 5, x = -5$ because that is where the denominator will be zero, and the derivatives (first and second) undefined. So just check what is happening there $f(0)$, however, will be zero as well.

$$f''(x) = \frac{(-4x)\sqrt{25-x^2}-(25-2x^2)\large\frac{-x}{\sqrt{25-x^2}}}{25-x^2} $$ $$ = \frac{(-4x)(25-x^2)-(25-2x^2)(-x)}{(25-x^2)\sqrt{25-x^2}} $$ $$ = \frac{-75x+2x^3}{(25-x^2)^{\large\frac32}} $$ $$ = \frac{x(2x^2 - 75)}{(25-x^2)^{\large\frac32}} $$ $$ = \frac{2x(x^2 - \frac{75}{2})}{(25 - x^2)^{\large \frac 32}}$$

Now, when is $f''(x) = 0$? Be careful with $x = 5, x= -5$: you want to check the point, but note that neither the first nor the second derivative is defined there.

When $2x = 0,\; f''(x) = 0$, as is $f'$ and $f$. When $x^2 - 75/2 = 0, f''(x) = 0 \implies x \pm 5\sqrt{\frac 32} = 0 \implies $

But to confirm that that there is an inflection point at $x=0$, you'll want to check whether $f''(x)$ changes signs on there, from negative to positive, or positive to negative. It does, indeed change signs at $x = 0$: as $x \to 0$ from the left, $f''(x) > 0$, and when $x \to 0$ from the right, $f''(x) <0$. So there is, indeed, an inflection point at $(0, 0)$.

  • $\begingroup$ So would my second derivative be correct? Then I can set it to zero to find the inflection point. $\endgroup$ – Fernando Martinez Mar 17 '13 at 18:17
  • $\begingroup$ I have an idea I can find a common denominator in my first derivative and then take the second one and then set it to zero. $\endgroup$ – Fernando Martinez Mar 17 '13 at 18:22
  • $\begingroup$ I think that's the route to go, because there's a mistake in your current $f''(x)$. $\endgroup$ – Namaste Mar 17 '13 at 18:24
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    $\begingroup$ It isn't: I corrected! Thanks! There is one zero: $x = 0$, as $x^2\geq 0$, hence no solution to $2x^2 + 75x = 0$ $\endgroup$ – Namaste Mar 17 '13 at 19:12
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    $\begingroup$ Yes Fernando. Very good observation. (See, we are all prone to stupid arithmetic errors.) I adjusted the points to check for inflection, accordingly. $\endgroup$ – Namaste Mar 22 '13 at 19:44

You have miscalculated your second derivative. First, I'd rewrite $$f'(x)=\sqrt{25-x^2}+\frac{-x^2}{\sqrt{25-x^2}}=\frac{25-2x^2}{\sqrt{25-x^2}},$$ from which we see that $$\begin{align}f''(x) &= \cfrac{(-4x)\sqrt{25-x^2}-(25-2x^2)\frac{-x}{\sqrt{25-x^2}}}{25-x^2}\\ &= \cfrac{(-4x)(25-x^2)-(25-2x^2)(-x)}{(25-x^2)\sqrt{25-x^2}}\\ &= \frac{-75x-2x^3}{(25-x^2)^\frac32}\\ &= \frac{x(x-5)(x+5)}{(25-x^2)^\frac32}.\end{align}$$ At which point(s), if any, does $f''(x)$ change signs? Such points are the inflection points. Note in particular that the denominator of $f''(x)$ is never negative--and always positive where $f''(x)$ is defined. The problem, then, amounts to finding the odd-multiplicity zeroes of the numerator of $f''(x)$ that are in the domain of $f''(x)$.

  • $\begingroup$ Using chin rule would it not be -2x/square root(25-x^2)? instead of -x $\endgroup$ – Fernando Martinez Mar 17 '13 at 18:44
  • $\begingroup$ Don't forget the $\frac12$ from the power rule. That will cancel with the $2$. $\endgroup$ – Cameron Buie Mar 17 '13 at 18:45
  • $\begingroup$ yes you are right thanks! $\endgroup$ – Fernando Martinez Mar 17 '13 at 18:46

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