Mister Exam
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-
How to use it?
- Graphing y =:
- x^2+3x-10
-
(x^2+3)/(x-1)
- x^2-10x+27
- ((x+1)^2)/(x-2)
-
Identical expressions
- (one +sec(x))^(one / four)
- (1 plus sec(x)) to the power of (1 divide by 4)
- (one plus sec(x)) to the power of (one divide by four)
- (1+sec(x))(1/4)
- 1+secx1/4
- 1+secx^1/4
- (1+sec(x))^(1 divide by 4)
-
Similar expressions
- (1-sec(x))^(1/4)
Graphing y = (1+sec(x))^(1/4)
The solution
You have entered
[src]
4 ____________ f(x) = \/ 1 + sec(x)
$$f{\left(x \right)} = \sqrt[4]{\sec{\left(x \right)} + 1}$$
f = (sec(x) + 1)^(1/4)
The graph of the function
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
$$\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{4 \left(\sec{\left(x \right)} + 1\right)^{\frac{3}{4}}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$
The values of the extrema at the points:
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:
$$x_{1} = 0$$
The function has no maxima
Decreasing at intervals
$$\left[0, \infty\right)$$
Increasing at intervals
$$\left(-\infty, 0\right]$$
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(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
$$\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{4 \left(\sec{\left(x \right)} + 1\right)^{\frac{3}{4}}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 0$$
The values of the extrema at the points:
4 ___ (0, \/ 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:
$$x_{1} = 0$$
The function has no maxima
Decreasing at intervals
$$\left[0, \infty\right)$$
Increasing at intervals
$$\left(-\infty, 0\right]$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \lim_{x \to -\infty} \sqrt[4]{\sec{\left(x \right)} + 1}$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \lim_{x \to \infty} \sqrt[4]{\sec{\left(x \right)} + 1}$$
True
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = \lim_{x \to -\infty} \sqrt[4]{\sec{\left(x \right)} + 1}$$
True
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = \lim_{x \to \infty} \sqrt[4]{\sec{\left(x \right)} + 1}$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (1 + sec(x))^(1/4), divided by x at x->+oo and x ->-oo
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \lim_{x \to -\infty}\left(\frac{\sqrt[4]{\sec{\left(x \right)} + 1}}{x}\right)$$
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x \lim_{x \to \infty}\left(\frac{\sqrt[4]{\sec{\left(x \right)} + 1}}{x}\right)$$
True
Let's take the limit
so,
inclined asymptote equation on the left:
$$y = x \lim_{x \to -\infty}\left(\frac{\sqrt[4]{\sec{\left(x \right)} + 1}}{x}\right)$$
True
Let's take the limit
so,
inclined asymptote equation on the right:
$$y = x \lim_{x \to \infty}\left(\frac{\sqrt[4]{\sec{\left(x \right)} + 1}}{x}\right)$$
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:
$$\sqrt[4]{\sec{\left(x \right)} + 1} = \sqrt[4]{\sec{\left(x \right)} + 1}$$
- Yes
$$\sqrt[4]{\sec{\left(x \right)} + 1} = - \sqrt[4]{\sec{\left(x \right)} + 1}$$
- No
so, the function
is
even
So, check:
$$\sqrt[4]{\sec{\left(x \right)} + 1} = \sqrt[4]{\sec{\left(x \right)} + 1}$$
- Yes
$$\sqrt[4]{\sec{\left(x \right)} + 1} = - \sqrt[4]{\sec{\left(x \right)} + 1}$$
- No
so, the function
is
even