Calculas
HW 10
1.
D(f) = \{x | -\infty < x < \infty \}\\
f'\left(x\right)=\frac{4}{5}x^{-\frac{1}{5}}\left(x-4\right)^2+x^{\frac{4}{5}}2\left(x-4\right)\\
f'\left(x\right)=x^{-\frac{1}{5}}\left(x-4\right)\left[\frac{4}{5}\left(x-4\right)+2x\right]\\
f'\left(x\right)=\frac{\left(x-4\right)\left[4\left(x-4\right)+10x\right]}{5\sqrt[5]{x}}\\
f'(x) = 0\ \text{or}\ f'(x)\ \text{is undefined when}\ x = \frac{8}{7}, x=4, x=0\\
\text{All critical points are in}\ f(x)\ \text{domain.}
2.
D(f) = \{x | -\infty < x < \infty \}\\
f'(x)=\frac{\frac{1}{3}\left(2x-1\right)^{-\frac{2}{3}}2\left(x-1\right)^2-\left(2x-1\right)^{\frac{1}{3}}2\left(x-1\right)}{\left(x-1\right)^4}\\
f'(x)=\frac{2\left(2x-1\right)^{-\frac{2}{3}}\left(x-1\right)\left[\frac{1}{3}\left(x-1\right)-2\left(2x-1\right)\right]}{\left(x-1\right)^4}\\
\frac{2\left(2x-1\right)^{-\frac{2}{3}}\left(x-1\right)\:\left[\:\frac{1}{3}\left(x-1\right)-\left(2x-1\right)\right]\:}{\left(x-1\right)^4}\\
f'(x)=\frac{2\:\left[\left(x-1\right)-3\left(2x-1\right)\right]\:}{3\sqrt[3]{\left(2x-1\right)^2}\left(x-1\right)^3}\\
f'(x)=\frac{2\:\left[-5x+2\right]\:}{3\sqrt[3]{\left(2x-1\right)^2}\left(x-1\right)^3}\\
f'(x) = 0\ \text{or}\ f'(x)\ \text{is undefined when}\ x = \frac{2}{5}, x=\frac{1}{2}, x=1\\
\text{All critical points are in}\ f(x)\ \text{domain.}
https://calcworkshop.com/application-derivatives/absolute-extrema/
HW 11
- Intermediate Value Theorem(I.V.T.)
- Rolle Theorem
- Mean Value Theorem
6
| x | 0 | 1 | 2 | 3 | 4 | ||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| f(x) | 6 | 4 | 2 | 8 | 12 | ||||||
| f'(x) | + | 0 | - | - | - | 0 | + | + | + | 0 | - |
| f''(x) | - | - | - | + | + | + | - | - | - |
7
Fomulas
f(x) = \sqrt{x^2 - 4}\\
D(f) = \{ x\ |\ x^2 -4 \geq 0\}\\
D(f) = \{ x\ |\ x \leq -2, x \geq 2 \}
ln rules
- Product Rule: ln (xy) = ln x + ln y
- Quotient Rule: ln (x/y) = ln x - lny
- Power Rule: ln x^y = y ln x
- Reciprocal Rule: ln (1/x) = -ln x
ln(1) = 0\\
ln(\infty)= \infty
TODO
- Limit and infinity
- sin cos tan rules
Edited by mon-jai





