User:Mmmooonnnsssttteeerrr/Calculus

This is the prototype page for Elementary Calculus. Contents written here may be transferred to there when a section is written here and if its content is deemed of acceptable standard for Elementary Calculus book.

Function
Function generates only one value for each value of an element. Function can be tested using Vertical Line Test.

Let's try to express the area of an equilateral triangle as a function of one of its sides. This problem can be solved by using property of 30-60-90 special right triangle. We know that 30-60-90 right triangle has side ratio of $$1:\sqrt{3}:2$$. Also, we know that we can divide an equilateral triangle into two 30-60-90 right triangles. Let x denote the base of an equilateral triangle and A(x) be the area of an equilateral triangle. Then,
 * $$A(x)=\frac{\frac{x}{2}\times\frac{\sqrt{3}}{2}}{2}\times2\,$$
 * $$=\frac{\sqrt{3}\times x}{4}\;, x>0\,$$

Functions can be classified into power function, root function, a degree of polynomial, rational function, algebraic function, trigonometric function, exponential function, or trigonometric function. Let's try to classify some functions ourselves. $$f(x)=\sqrt[5]{x}$$ is a root function. $$g(x)=\sqrt{1-x^2}$$ is an algebraic function. $$h(x)=x^9+x^4$$ is a polynomial of degree 9. $$r(x)=\frac{x^2+1}{x^3+x}$$ is a rational function. $$s(s)=\tan 2x$$ is a trigonometric function. $$t(x)=\log_{10} x$$ is a logarithmic function. Transcendental functions include trigonometric, inverse trigonometric, exponential, and logarithmic functions.

In statistical study of linear regression, there is a way to conceptualize function as joint distribution of marginal distribution P(x) of explanatory variable x and conditional distribution P(y|x) of response variable y. Explanatory variable is also called endogenous variable or input variable. Response variable is also called exogenous variable or output variable. Of course, there can be more than one explanatory variables in marginal distribution and more than one response variables in conditional distribution.

Tutorial 2
1. (a) Sketch $$f(x)=\sqrt{x}-4$$ and state its domain and range.

To find the domain we know that x>0, so the domain is x>0. To find the range we consider x,y-intercepts, domain, and shape of the curve. The x-intercept is 16, the y-intercept is -4, we know the domain is x>0, and we know that f(x) is $$\sqrt{x}$$ moved downwards by 4 units. Therefore, the range for f(x) is y>-4. $$\Box$$

(b) $$f(f^{-1}(x))=x$$, use this to find the inverse function of f(x) and state its domain and range.

$$x=f(f^{-1}(x))=\sqrt{f^{-1}(x)}-4$$

$$x+4=\sqrt{f^{-1}(x)}$$

$$(x+4)^2=f^{-1}(x)$$

So, the inverse function of f(x) is $$(x+4)^2$$. By the definition of inverse function, we know that the domain of the inverse function is the range of f(x), and the range of the inverse function is the domain of f(x). Therefore, the domain of the inverse function is x>-4, and the range is y>0.


 * composite functions of trignometric and inverse trigonometric functions

Theorem
$$\int_1^\infty\frac{1}{x^p}dx=\begin{cases}p>1,integral\;converges\\p\le1,integral\;diverges\end{cases}$$

My Proof
$$\int_{1}^{\infty}\frac{1}{x^p}dx=lim_{a\rightarrow\infty}\int_1^a\frac{1}{x^p}dx=\begin{cases}lim_{a\rightarrow\infty}\left[\frac{x^{-p+1}}{-p+1}\right]^a_1,p\ne1\\lim_{a\rightarrow\infty}\left[ln x^p\right]^a_1,p=1\end{cases}$$

If $$p>1, -p+1<0$$:

Let $$|-p+1|=k$$,

$$\left[\frac{x^{-k}}{-k}\right]^a_1=\frac{a^{-k}}{-k}-\frac{1}{-k}=\frac{1}{-ka^k}-\frac{1}{-k}$$

Therefore, $$lim_{a\rightarrow\infty}\left[\frac{1}{-ka^k}+\frac{1}{k}\right]=\frac{1}{k}.$$

If $$p<1, -p+1>0$$:

Let $$-p+1=k$$

$$\left[\frac{x^k}{k}\right]^a_1=\frac{a^k}{k}-\frac{1}{k}$$

Therefore, $$lim_{a\rightarrow\infty}\left[\frac{a^k}{k}-\frac{1}{k}\right]\rightarrow\infty.$$

If $$p=1$$:

$$\left[ln x^p\right]^a_1=ln a^p-ln 1=ln a^p$$

Therefore, $$lim_{a\rightarrow\infty}\left[ln a^p\right]\rightarrow\infty.\Box$$