Mister Exam

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  • Graphing y =:
  • |x^2+x-2| |x^2+x-2|
  • x(2-x)^2
  • x^2-6x+9
  • -x^2+6x-3
  • Identical expressions

  • one /(cos(x)+ one / two)
  • 1 divide by ( co sinus of e of (x) plus 1 divide by 2)
  • one divide by ( co sinus of e of (x) plus one divide by two)
  • 1/cosx+1/2
  • 1 divide by (cos(x)+1 divide by 2)
  • Similar expressions

  • 1/(cos(x)-1/2)
  • 1/(cosx+1/2)

Graphing y = 1/(cos(x)+1/2)

v

The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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            1      
f(x) = ------------
       cos(x) + 1/2
$$f{\left(x \right)} = \frac{1}{\cos{\left(x \right)} + \frac{1}{2}}$$
f = 1/(cos(x) + 1/2)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 2.0943951023932$$
$$x_{2} = 4.18879020478639$$
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0
so we need to solve the equation:
$$\frac{1}{\cos{\left(x \right)} + \frac{1}{2}} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to 1/(cos(x) + 1/2).
$$\frac{1}{\frac{1}{2} + \cos{\left(0 \right)}}$$
The result:
$$f{\left(0 \right)} = \frac{2}{3}$$
The point:
(0, 2/3)
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
$$\frac{\sin{\left(x \right)}}{\left(\cos{\left(x \right)} + \frac{1}{2}\right)^{2}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$
$$x_{2} = \pi$$
The values of the extrema at the points:
(0, 2/3)

(pi, -2)


Intervals of increase and decrease of the function:
Let's find intervals where the function increases and decreases, as well as minima and maxima of the function, for this let's look how the function behaves itself in the extremas and at the slightest deviation from:
Minima of the function at points:
$$x_{1} = 0$$
Maxima of the function at points:
$$x_{1} = \pi$$
Decreasing at intervals
$$\left[0, \pi\right]$$
Increasing at intervals
$$\left(-\infty, 0\right] \cup \left[\pi, \infty\right)$$
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
$$\frac{4 \left(\cos{\left(x \right)} + \frac{4 \sin^{2}{\left(x \right)}}{2 \cos{\left(x \right)} + 1}\right)}{\left(2 \cos{\left(x \right)} + 1\right)^{2}} = 0$$
Solve this equation
Solutions are not found,
maybe, the function has no inflections
Vertical asymptotes
Have:
$$x_{1} = 2.0943951023932$$
$$x_{2} = 4.18879020478639$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty} \frac{1}{\cos{\left(x \right)} + \frac{1}{2}} = \left\langle -\infty, \infty\right\rangle$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \left\langle -\infty, \infty\right\rangle$$
$$\lim_{x \to \infty} \frac{1}{\cos{\left(x \right)} + \frac{1}{2}} = \left\langle -\infty, \infty\right\rangle$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \left\langle -\infty, \infty\right\rangle$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of 1/(cos(x) + 1/2), divided by x at x->+oo and x ->-oo
True

Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \lim_{x \to -\infty}\left(\frac{1}{x \left(\cos{\left(x \right)} + \frac{1}{2}\right)}\right)$$
True

Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x \lim_{x \to \infty}\left(\frac{1}{x \left(\cos{\left(x \right)} + \frac{1}{2}\right)}\right)$$
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:
$$\frac{1}{\cos{\left(x \right)} + \frac{1}{2}} = \frac{1}{\cos{\left(x \right)} + \frac{1}{2}}$$
- Yes
$$\frac{1}{\cos{\left(x \right)} + \frac{1}{2}} = - \frac{1}{\cos{\left(x \right)} + \frac{1}{2}}$$
- No
so, the function
is
even