calculusderivatives
Can anyone help me find the points of inflection in the following function in the interval between $0$ and $2\pi$
$f(x)=\sqrt{2}x^2-4\sin(x)$
for my first derivative I got
$f'(x)=2\sqrt{2}x-4\cos(x)$
$f''(x)=2\sqrt{2}+4\sin(x)$
$\sin(x)=\frac{-\sqrt{2}}{2}$
inflection points would be $x=\frac{5\pi}{4},\frac{7\pi}{4}$
but would this be correct
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 $(0, 0)$.
For any $x\in [0,2\pi]$ you get $f'(x)=\cos (x) -\sin (x)$ and $f''(x)=-\sin (x)-\cos (x)$. Therefore $$\begin{align} f'(x)=0&\iff \cos (x) -\sin(x)=0\\ &\iff \cos (x) =\sin (x)\\ &\iff x\in \{\frac{\pi}{4}, \frac{5\pi}{4}\} \end{align}$$
Which gives you the critical points $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{5\pi}{4}$.
The last equivalence is easy to see geometrically if you look at the unit circle.
I don't understand why you're messing with $\cos ^{-1}$. Care to explain so we can help you?
Similarly, for $f''$ you'll get the zeros $\displaystyle \frac{3\pi}{4}$ and $\displaystyle \frac{7\pi}{4}$.
Best Answer
That is correct, but one more remark is worth making:
An inflection point is not merely a point where the second derivative is $0$, but rather is a point where the second derivative changes from positive to negative or vice-versa.
For example, if $g(x)=x^4$, then $g''(x)=0$ when $x=0$, but that's not an inflection point since $g''(x)$ is positive if $x$ is on either side of $0$.