Answers to Calculus Practical 1

Question 1 §

The domain and range of the functions in question 1 are as follows, respectively:

$$\text{Domain: }\mathbb{R},\text{Range: }[0,\infty )$$

$$\text{Domain: }\mathbb{R},\text{Range: }\mathbb{R}$$

$$\text{Domain: }[-\sqrt{5},\sqrt{5}],\text{Range: }[0,\sqrt{5}]$$

$$\text{Domain: }[0,\infty ),\text{Range: }[0,\infty )$$

$$\text{Domain: }(-\infty ,0)\cup (0,\infty ),\text{Range: }(-\infty ,0)\cup (0,\infty )$$

$$\text{Domain: }(-\infty ,0)\cup (0,\infty ),\text{Range: }(0,\infty )$$

$$\text{Domain: }(-\infty ,3)\cup (3,\infty ),\text{Range: }(-\infty ,0)\cup (0,\infty )$$

$$\text{Domain: }(-2,2),\text{Range: }[\frac{1}{2},\infty )$$

$$\text{Domain: }(-\infty ,a],\text{Range: }[0,\infty )$$

$$\text{Domain: }(-\infty ,b)\cup (b,\infty ),\text{Range: }(0,\infty )$$

Question 2 §

The degree of the respective polynomials shown in the question would be: 1, 3, 7, 0.

Question 3 §

Complete at home with my drawing tablet Cos graph.excalidraw

Question 4 §

Complete at home with my drawing tablet

Question 5 §

Complete at home with my drawing tablet Yes, in the lecture we proposed splitting the function into two functions for the top and the bottom - thus here we can split the function in the two sides: $x^2+y^2=r^2⟹$

Question 6 §

The function would be equal to $\frac{0}{0}=\text{undefined}$, thus $x=-2$ shouldn’t be included in the domain.

Question 7 §

The function of $\cos$ as a power series is as follows:

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$$

Differentiating that gives the function of $-\sin$ as shown below:

$$\frac{d}{dx}(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots )=-x+\frac{x^3}{3!}-\dots =-(x-\frac{x^3}{3!}+\dots )=-\sin x$$

Question 8 §

${\sin }^{2}x+{\cos }^{2}x\equiv 1⟹\frac{{\sin }^{2}x}{{\sin }^{2}x}+\frac{{\cos }^2}{{\sin }^{2}x}x\equiv \frac{1}{{\sin }^{2}x}⟹1+{\cot }^{2}x\equiv {\csc }^{2}x$ Which is equivalent to the required result.

To show that $\frac{2\sin x}{1-\cos x}-\frac{2\cos x}{\sin x}\equiv 2\csc x$, it is a bit more involved: ${\sin }^{2}x+{\cos }^{2}x\equiv 1$ $⟹2{\sin }^{2}x+2{\cos }^{2}x\equiv 2$ $⟹2{\sin }^{2}x-2\cos x+2{\cos }^{2}x+2\cos x\equiv 2$ $⟹2{\sin }^{2}x-(2\cos x)(1-\cos x)\equiv 2-2\cos x$ $⟹\frac{2{\sin }^{2}x}{1-\cos x}-2\cos x\equiv 2$

Question 9 §

As $n$ increases, the gradient change between the two asymptotes is quicker, slowly tending towards perfectly meeting the asymptotes looking like a right-angle. This is explainable through the derivative of the function (which represents the gradient) similarly growing in order (thus gradient changes occur exponentially quickly), so by about $n=11$ it already looks like a right angle.

Question 10 §

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n}}{(2n)!}$$

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots =\sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}$$