Attribution-ShareAlike 4.0 International

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The set of value taken by $f(x)$ when $x$ takes all possible value in the domain is the range of $f(x)$.

where $f$ and $g$ are polynomials, is called a rational function.

Its range has to exclude all those values of $x$ where $g(x) = 0$.

Limits from Above and Below

Sometimes approaching 0 with small positive values of $x$ gives you a different limit from approaching with small negative values of $x$.

The limit you get from approaching 0 with positive values is known as the limit from above:

$$\lim_{x \rightarrow a^+} f(x)$$

and with negative values is known as the limit from below:

$$\lim_{x \rightarrow a^-} f(x)$$

If the two limits are equal, we simply refer to the limit.

It can also be written as an infinite series:

$$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$

The two important limits to know are:

• as $x \rightarrow + \infty$, $\exp x \rightarrow +\infty$ ($e^x \rightarrow +\infty$)
• as $x \rightarrow -\infty$, $\exp x \rightarrow 0$ ($e^x \rightarrow 0$)

Note that $e^x > 0$ for all real values of $x$.

Some key facts about these functions:

• $\cosh$ has even symmetry and $\sinh$ and $\tanh$ have odd symmetry
• as $x \rightarrow + \infty$, $\cosh x \rightarrow +\infty$ and $\sinh x \rightarrow +\infty$
• $\cosh^2x - \sinh^2x = 1$
• $\tanh$’s limits are -1 and +1
• Derivatives:
  • $\frac{\mathrm{d}}{\mathrm{d}x} \sinh x = \cosh x$
  • $\frac{\mathrm{d}}{\mathrm{d}x} \cosh x = \sinh x$
  • $\frac{\mathrm{d}}{\mathrm{d}x} \tanh x = \frac{1}{\cosh^2x}$

Since the exponential of any real number is positive, the domain of $\ln$ is $x > 0$.

To draw the curve of an implicit function you have to rewrite it in the form $y = f(x)$. There may be more than one $y$ value for each $x$ value.

As $h\rightarrow 0$, the clospe of the cord $\rightarrow$ slope of the tangent, or:

$$f'(x_0) = \lim_{h\rightarrow0}\frac{f(x_0+h) - f(x_0)}{h}$$

whenever this limit exists.

Let $u(x) = x^2$, $F(u) = \cos u$

$$\frac{\mathrm df}{\mathrm dx} = -\sin u \cdot 2x = 2x\sin{x^2}$$

1. Set $f(0) = g(0$:

   $$f(0) = 1 \rightarrow g(0) = a_0 = 1$$

2. Set $f'(0) = g'(0$:

   $$f'(0) = -\sin0 = 0 \rightarrow g'(0) = a_1 = 0$$

3. Set $f''(0) = g''(0$:

   $$f''(0) = -\cos = -1 \rightarrow g''(0) = 2a_2 = -1 \rightarrow a_2 = -0.5$$

So for $x$ near 0,

$$\cos x \approx 1 - \frac 1 2 x^2$$

Check:

1. Find the crossing points $P$ and $Q$

   $$\begin{align*} f_1(x) &= f_2(x) \\ x &= 2-x^2 \\ x &= 1 \\ x &= -2 \end{align*}$$

2. Since $f_1(x) \ge f_2(x)$ between $P$ and $Q$

$$\begin{align*} A &= \int^1_{-2}\! (f_1(x) - f_2(x)) \mathrm dx \\ &= \int^1_{-2}\! (2 - x^2 - x) \mathrm dx \\ &= \left[ 2x - \frac 13 x^3 - \frac 12 x^2 \right]^1_{-2} \\ &= \left(2 - \frac 13 - \frac 12 \right) - \left( -4 + \frac 83 - \frac 42 \right) \\ &= \frac 92 \end{align*}$$

1. Let $w(x) = 5x - 1$

2. $$\begin{align*} \frac{\mathrm d}{\mathrm dx} w &= 5 \\ \frac 15 \mathrm dw &= \mathrm dx \end{align*}$$

3. The integral is then

   $$\begin{align*} I &= \int\! w^3 \frac 15 \mathrm dw \\ &= \frac 15 \cdot \frac 14 \cdot w^4 + c \\ &= \frac{1}{20}w^4 + c \end{align*}$$

4. Finally substitute $w$ out

$$I = \frac{(5x-1)^4}{20} + c$$

1. Let

   $$w(x) = \sin x + 1$$

2. Then

   $$\begin{align*} \frac{\mathrm d}{\mathrm dx} w = \cos x \\ \mathrm dw = \cos x \mathrm dx \\ \end{align*}$$

3. The integral is now

   $$\begin{align*} I &= \int\! \sqrt w \,\mathrm dw \\ &= \int\! w^{\frac12} \,\mathrm dw \\ &= \frac23w^{\frac32} + c \end{align*}$$

4. Finally substitute $w$ out to get:

   $$I = \frac23 (\sin x + 1)^{\frac32} + c$$

1. Use the previous example to get to

   $$I = \int^2_1\! \sqrt w \,\mathrm dw = \frac23w^{\frac32} + c$$

2. Since $w(x) = \sin x + 1$ the limits are:

   $$\begin{align*} x = 0 &\rightarrow w = 1\\ x = \frac\pi2 &\rightarrow w = 2 \end{align*}$$

3. This gives us

   $$I = \left[ \frac23w^{\frac32} \right]^2_1 = \frac23 (2^{\frac23} = 1)$$

1. Try a trigonmetrical substitution:

$$\begin{align*} x &= \sin w \\ \\ \frac{\mathrm dx}{\mathrm dw} = \cos w \\ \mathrm dx = \cos 2 \,\mathrm dw \\ \end{align*}$$

2. $$\begin{align*} x=0 &\rightarrow w=0 \\ x=1 &\rightarrow w=\frac\pi2 \end{align*}$$

3. Therefore

   $$\begin{align*} I &= \int^{\frac\pi2}_0\! \sqrt{1 - \sin^2 w} \cos w \,\mathrm dw \\ &= \int^{\frac\pi2}_0\! \cos^w w \,\mathrm dw \end{align*}$$

   But $\cos(2w) = 2\cos^2w - 1$ so:

   $$\cos^2w = \frac12 \cos(2w) + \frac12$$

   Hence

   $$\begin{align*} I &= \int^{\frac\pi2}_0\! \frac12 \cos(2w) + \frac12 \,\mathrm dw \\ &= \left[ \frac14 \sin(2w) + \frac w2 \right]^{\frac\pi2}_0 \\ &= \left( \frac14 \sin\pi + \frac\pi4 \right) - 0 \\ &= \frac\pi4 \end{align*}$$

Integration by Parts

$$uv = \int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx + \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$

or

$$\int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx = uv - \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$

This technique is derived from integrating the product rule.

1. Use

   $$\int\! u\frac{\mathrm dv}{\mathrm dx} \,\mathrm dx = uv - \int\! \frac{\mathrm du}{\mathrm dx}v \,\mathrm dx$$

2. Set $u = \ln x$
   and $v' = 1$.

3. This means that $u' = \frac1x$ and $v = x$.

4. $$\begin{align*} I &= x\ln x - \int\! x\cdot\frac1x \,\mathrm dx + c \\ &= x\ln x - \int\! \,\mathrm dx + c \\ &= x\ln x - x + c \\ \end{align*}$$

Rearrange to get

$$\begin{align*} \int\! \frac1y \,\mathrm dy &= \int\! k \mathrm dx + c \\ \ln y &= kx + c \\ y &= e^{kx + c} = e^ce^{kx} \\ &= Ae^{kx} \end{align*}$$

where $A = e^c$ is an arbitrary constant.