Mister Exam

Graphing y = cos4x+sin8x

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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f(x) = cos(4*x) + sin(8*x)
$$f{\left(x \right)} = \sin{\left(8 x \right)} + \cos{\left(4 x \right)}$$
f = sin(8*x) + cos(4*x)
The graph of the function
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to cos(4*x) + sin(8*x).
$$\sin{\left(0 \cdot 8 \right)} + \cos{\left(0 \cdot 4 \right)}$$
The result:
$$f{\left(0 \right)} = 1$$
The point:
(0, 1)
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$- 4 \sin{\left(4 x \right)} + 8 \cos{\left(8 x \right)} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$- 16 \left(4 \sin{\left(8 x \right)} + \cos{\left(4 x \right)}\right) = 0$$
Solve this equation
The roots of this equation
$$x_{1} = - \frac{7 \pi}{8}$$
$$x_{2} = - \frac{3 \pi}{8}$$
$$x_{3} = \frac{\pi}{8}$$
$$x_{4} = \frac{5 \pi}{8}$$

Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Concave at the intervals
$$\left[\frac{5 \pi}{8}, \infty\right)$$
Convex at the intervals
$$\left(-\infty, - \frac{7 \pi}{8}\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\sin{\left(8 x \right)} + \cos{\left(4 x \right)}\right) = \left\langle -2, 2\right\rangle$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \left\langle -2, 2\right\rangle$$
$$\lim_{x \to \infty}\left(\sin{\left(8 x \right)} + \cos{\left(4 x \right)}\right) = \left\langle -2, 2\right\rangle$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \left\langle -2, 2\right\rangle$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of cos(4*x) + sin(8*x), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\sin{\left(8 x \right)} + \cos{\left(4 x \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\sin{\left(8 x \right)} + \cos{\left(4 x \right)}}{x}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x).
So, check:
$$\sin{\left(8 x \right)} + \cos{\left(4 x \right)} = - \sin{\left(8 x \right)} + \cos{\left(4 x \right)}$$
- No
$$\sin{\left(8 x \right)} + \cos{\left(4 x \right)} = \sin{\left(8 x \right)} - \cos{\left(4 x \right)}$$
- No
so, the function
not is
neither even, nor odd