Extrema of $f(x)=\frac{\sin (5x)} 5 - \frac{2\sin(3x)} {3} + \sin (x)$.

Question

(a) I need help in finding maxima and minima of the following funcion:

$$f(x)=\frac{\sin (5x)} 5 - \frac{2\sin(3x)} {3} + \sin (x)$$

therefore I need to find the roots of $f'(x)=\cos(5x)-2\cos(3x)+\cos(x)$.

(b) I need to find the minima of $$f(x)=4x+\frac{9\pi^2} x + \sin x.$$ How can I find the roots of $$f'(x)=4-\frac{9\pi^2 }{x^2} + \cos x?$$

Thank you.

Answer 1

Hints:

a) $\cos(5x) \equiv 16\cos^5(x)-20\cos^3(x)+5\cos(x)$

$\cos(3x) \equiv 4\cos^3(x)-3\cos(x)$

So, we've got:

Now, let $c \equiv \cos(x)$.

We've got:

$$[16c^5-10c^3+5c]-2[4c^3-3c]+c=0 \iff \boxed{2c\underbrace{[8c^4-9c^2+6=0]}_{\textrm{quadratic in} \ c^2}} \tag{A}$$

Solve $(A)$!

b) Solving this analytically is very hard, so use the bisection method (noting that $4-\frac{9\pi^2}{x^2}=\cos(x)=0$ is continuous on $\mathbb{R \setminus\{0\}}$).

The exact answer to b) is $\boxed{\pm \frac{3\pi}{2}}$.

Now, your job is to evaluate the second derivative of these extrema, in order to determine their nature (maxima or minima).

Answer 2

Hint for (a): $\cos 5x+\cos x=2\cos 3x\cos 2x$

For (b), I think you will need a numerical method, such as Newton's Method.

Answer 3

$$\begin{align} \cos5x-2\cos3x+\cos x&=\cos5x-\cos 3x + \cos x-\cos 3x\\ &=-2\sin4x\sin x+2\sin2x\sin x\\ &=2\sin x(-\sin 4x+\sin2x)\\ &=-4\sin x\sin3x\cos x \end{align}$$

Now you only have to equal $0$ to each factor.