Finding an inflection point 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 
$f(x)=x\sqrt{25-x^2}$
for my first derivative I did
$f'(x)=1\sqrt{25-x^2}+(x)\frac{1}{2\sqrt{25-x^2}}(-2x)$
for my common denominator I got
$\frac{50-4x^2}{2\sqrt{25-x^2}}$
for my second derivative I got
$f''(x)=\frac{(-4x)\sqrt{25-x^2}-(25-2x^2)(1/2)\sqrt{25-x^2}(-2x)}{\sqrt{(25-x^2)^2}}$
But I am having trobule finding the inflection point.
 A: 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 $(0, 0)$.
A: 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 $f''(x)$.
