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.
Best Answer
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 $(0, 0)$.