The course of the function $f(x) = \sin^3 (x) + \cos^3 (x)$.

Question

First, sorry for my english, hope you understand.

so I'm examining the course of this function $f(x) = \sin^3 (x) + \cos^3 (x)$ where I need to find everything I can about this function (e.g. definition field, range of values, inflection points, asymptots, global and local max/min, etc.). Now I'm struggling with 3 tasks.

1. Prove there are no limits in $+\infty$ by showing there are different limits of $f(\pi/2+2k\pi)$ and $f(2k\pi)$ where $k$ goes to $+\infty$. Is this right? Are the results of the limits different?

2. How to get inflection points? I know I get them by solving the 2nd derivative of $f(x)$ which I think I have but I don't know how to get the exact values. graph and inflex points

3. Find intersection of $f(x)$ with axes. I got to the point where I have two equations and I need to prove why is it like that.

Thank you in advance

edit: Thank you all for helping me with this problem. I've used @Robert Lee solution in the end.

Answer 1

COMMENT.-(1)Any periodic function defined over $\mathbb R$ has a limit when $x\to\infty$$\space (f$ non-constant).

(2) $f(x)=(\sin(x)+\cos(x))(1-\sin(x)\cos(x))$ so the first factor gives the zeros of the function $x=-\dfrac{\pi}{4}+k\pi$ where $k\in\mathbb Z$. (the second factor clearly has
non-real roots).

(3) Inflection points: you have $f''(x)=-3(\cos(x)-\sin(x))^2(\cos(x)+\sin(x))=0$ find yourself $f'''(x)$ and determine this thirds question

Answer 2

(1) the idea is right, note that $\sin(k\pi) = 0$ and $\cos(k\pi) = (-1)^k$

(2) Note $$ \begin{split} f'(x) &= 3\sin^2 x \cos x - 3\cos^2x\sin x = 3 \sin x \cos x (\sin x - \cos x)\\ f''(x) &= \ldots = -3 (\cos x - \sin x)^2(\cos x + \sin x)\\ \end{split} $$ and now solving $f''(x)=0$ is trivial.

(3) your argument makes sense, if you are asking something else, please clarify it

Answer 3

1)

Notice that if $k \in \mathbb{Z}$ we have that $$ \sin\left( 2k\pi - \frac{\pi}{2} \right) = -\cos\left(2k\pi \right)= -1 $$ and $$ \cos\left( 2k\pi - \frac{\pi}{2} \right) = \sin\left(2k\pi \right)= 0 $$ for the first limit. For the second, notice that $$ \sin\left( 2k\pi \right) = 0 $$ and $$ \cos\left( 2k\pi \right) =1 $$ Finally, notice that $$ \lim_{k \to \infty} (-1)^3 + 0^3 = -1 \color{red}{\neq} 1 = \lim_{k \to \infty} (0)^3 + 1^3 $$

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2)

This one's a bit tricky. By repeated application of the chain rule and then using trigonometric identities we get the following: \begin{align*} \frac{d^2}{dx^2}\sin^3(x) + \cos^3(x) &= 3 \left[2 \sin(x) \cos^2(x) - \underbrace{\sin^3(x)}_{\color{blue}{\left(1-\cos^2(x)\right) \sin(x)}} + 2 \cos(x) \sin^2(x) - \underbrace{\cos^3(x)}_{\color{blue}{\left(1-\sin^2(x)\right) \cos(x)}}\right] \\ & = 3 \left[2 \sin(x) \cos^2(x) \color{blue}{-\sin(x) + \cos^2(x) \sin(x)} + 2 \cos(x) \sin^2(x) \color{blue}{-\cos(x) + \sin^2(x) \cos(x)}\right]\\ & = 3 \left[\color{blue}{3} \sin(x) \cos^2(x) + \color{blue}{3} \cos(x) \sin^2(x) - \sin(x) - \cos(x)\right]\\ & =\frac{3}{2} \left[6 \sin(x) \cos^2(x) + 6 \cos(x) \sin^2(x) -2 \sin(x) - 2\cos(x)\right]\\ & =\frac{3}{\sqrt{2}\sqrt{2}}\left( \sin(x) + \cos(x)\right)\left(6 \ \underbrace{\sin(x) \cos(x)}_{\color{purple}{\frac{\sin(2x)}{2}}} -2 \right)\\ & =\frac{3}{\sqrt{2}}\left( \sin(x)\underbrace{\frac{1}{\sqrt{2}}}_{\color{green}{\cos\left(\frac{\pi}{4}\right)}} + \cos(x)\underbrace{\frac{1}{\sqrt{2}}}_{\color{green}{\sin\left(\frac{\pi}{4}\right)}}\right)\left(\color{purple}{3}\sin(2x) -2 \right)\\ & = \frac{3}{\sqrt{2}}\sin\left(x + \frac{\pi}{4}\right)\left(3\sin(2x) -2 \right) \end{align*} We can set this last equation equal to $0$, which would them imply that each of the factors is equal to $0$ separately. Recalling that $$ \sin(\theta) = \alpha \color{blue}{\iff} \theta = (-1)^n\arcsin\left(\alpha\right) + \pi n, \quad n \in \mathbb{Z} \tag{1} $$ We then see that the inflection points are at \begin{align} \sin\left(x + \frac{\pi}{4}\right) = 0 &\iff x + \frac{\pi}{4} = (-1)^n\arcsin\left(0\right) + \pi n\\ &\iff x + \frac{\pi}{4} = \pi n\\ &\iff \boxed{x = \pi n - \frac{\pi}{4}}, \quad n \in \mathbb{Z} \end{align} And also at \begin{align} 3\sin(2x) -2 = 0 & \iff \sin(2x) = \frac{2}{3}\\ &\iff 2x = (-1)^n\arcsin\left(\frac{2}{3}\right) + \pi n\\ &\iff x = \boxed{\frac{(-1)^n}{2}\arcsin\left(\frac{2}{3}\right) + \frac{\pi}{2} n},\quad n \in \mathbb{Z} \end{align}

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3)

Your solution is correct. The zeros of the function are at the points where $$ \sin(x) = - \cos(x) \color{blue}{\implies}\tan(x) = -1 $$ And recalling that $$ \tan(\theta) = \alpha \color{blue}{\iff} \theta = \arctan\left(\alpha\right) + \pi n, \quad n \in \mathbb{Z} \tag{2} $$ We see that the zeros are at the points \begin{align} \tan(x) = -1 & \iff x = \arctan(-1) + \pi n\\ & \iff x = \boxed{-\frac{\pi}{4} + \pi n}, \quad n \in \mathbb{Z} \end{align} Which is exactly the same set that you propose in your solution since, if we start counting the integers from one over by setting $$ n = k+1 \in \mathbb{Z} $$ we get \begin{align} x &=-\frac{\pi}{4} + \pi n\\ &=-\frac{\pi}{4} + \pi (k+1)\\ &=-\frac{\pi}{4} + \pi k+ \pi\\ &=\pi\left(1-\frac{1}{4}\right) + \pi k\\ &=\frac{3\pi}{4} + \pi k, \qquad k \in \mathbb{Z} \end{align}

Also, if you want to argue why $1 = \sin(x)\cos(x)$ has no solutions, recall the previously used identity that $\sin(x)\cos(x) = \frac{\sin(2x)}{2}$. So you want to show that $$ \frac{\sin(2x)}{2} = 1 \color{blue}{\implies} \sin(2x) = 2 $$ but recalling that the sine function can only output values in the interval $[-1,1]$, since $2 \notin [-1,1] $ the above equation has no solutions.

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If you have any questions about the solutions please let me know!