Mister Exam
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- Graph of implicit function
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- Graph of parametric function
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-
How to use it?
- Graphing y =:
-
x^3-12x
- x/(x^2+1)
- x^3+3x
- x^2(x-2)^2
-
Identical expressions
- (two π₯+ one)π^ three ^π₯^2β4π₯+ one
- (2π₯ plus 1)π cubed to the power of π₯ squared β4π₯ plus 1
- (two π₯ plus one)π to the power of three to the power of π₯ squared β4π₯ plus one
- (2π₯+1)π3π₯2β4π₯+1
- 2π₯+1π3π₯2β4π₯+1
- (2π₯+1)πΒ³^π₯Β²β4π₯+1
- (2π₯+1)π to the power of 3 to the power of π₯ to the power of 2β4π₯+1
- 2π₯+1π^3^π₯^2β4π₯+1
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Similar expressions
- (2π₯-1)π^3^π₯^2β4π₯+1
- (2π₯+1)π^3^π₯^2β4π₯-1
Graphing y = (2π₯+1)π^3^π₯^2β4π₯+1
The solution
You have entered
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f(x) = (2*x + 1)*E - 4*x + 1
$$f{\left(x \right)} = \left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1$$
f = E^(3^(x^2))*(2*x + 1) - 4*x + 1
Horizontal asymptotes
Letβs find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesnβt exist
$$\lim_{x \to \infty}\left(\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesnβt exist
$$\lim_{x \to -\infty}\left(\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1\right) = -\infty$$
Let's take the limit
so,
horizontal asymptote on the left doesnβt exist
$$\lim_{x \to \infty}\left(\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1\right) = \infty$$
Let's take the limit
so,
horizontal asymptote on the right doesnβt exist
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of (2*x + 1)*E^(3^(x^2)) - 4*x + 1, divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the left doesnβt exist
$$\lim_{x \to \infty}\left(\frac{\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesnβt exist
$$\lim_{x \to -\infty}\left(\frac{\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the left doesnβt exist
$$\lim_{x \to \infty}\left(\frac{\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1}{x}\right) = \infty$$
Let's take the limit
so,
inclined asymptote on the right doesnβt exist
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) ΠΈ f = -f(-x).
So, check:
$$\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1 = 4 x + \left(1 - 2 x\right) e^{3^{x^{2}}} + 1$$
- No
$$\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1 = - 4 x - \left(1 - 2 x\right) e^{3^{x^{2}}} - 1$$
- No
so, the function
not is
neither even, nor odd
So, check:
$$\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1 = 4 x + \left(1 - 2 x\right) e^{3^{x^{2}}} + 1$$
- No
$$\left(e^{3^{x^{2}}} \left(2 x + 1\right) - 4 x\right) + 1 = - 4 x - \left(1 - 2 x\right) e^{3^{x^{2}}} - 1$$
- No
so, the function
not is
neither even, nor odd