Pengguna:S Rifqi/Bak pasir 1

Halaman ini hanyalah bak pasir. Halaman ini dapat berubah isi sewaktu-waktu.

1. Pertidaksamaan

a. ${\textstyle {\frac {2x+7}{|x-1|}}\geq 1}$

Syarat penyebut

${\begin{aligned}|x-1|&\neq 0\\x-1&\neq 0\\x&\neq 1\end{aligned}}$

Syarat nilai mutlak

$|x-1|=\left\{{\begin{array}{r}x-1&{\text{jika }}x\geq 1\\1-x&{\text{jika }}x<1\end{array}}\right.$

Pertidaksamaan

${\begin{aligned}{\frac {2x+7}{|x-1|}}&\geq 1\\{\frac {2x+7}{|x-1|}}-1&\geq 0\\{\frac {2x+7-|x-1|}{|x-1|}}&\geq 0\end{aligned}}$

Kasus I Untuk ${\textstyle x\geq 1}$,

${\begin{aligned}{\frac {2x+7-(x-1)}{(x-1)}}&\geq 0\\{\frac {2x+7-x+1}{x-1}}&\geq 0\\{\frac {x+8}{x-1}}&\geq 0\end{aligned}}$

Harga nol:

<-----+     +----->
  (+) | (-) | (+)
------●-----●------
     -8     1

Dengan menerapkan syarat-syarat yang berlaku, ${\textstyle {\text{HP}}=\{x>1\}}$.

Kasus II Untuk ${\textstyle x<1}$,

${\begin{aligned}{\frac {2x+7-(1-x)}{(1-x)}}&\geq 0\\{\frac {2x+7-1+x}{1-x}}&\geq 0\\{\frac {3x+6}{1-x}}&\geq 0\end{aligned}}$

Harga nol:

      +-----+
  (-) | (+) | (-)
------●-----●------
     -2     1

Dengan menerapkan syarat-syarat yang berlaku, ${\textstyle {\text{HP}}=\{-2\leq x<1\}}$.

Jadi, ${\text{HP}}$ setelah digabung adalah ${\textstyle \{-2\leq x<1{\text{ atau }}x>1\}}$ atau dalam notasi interval ${\textstyle [-2,1){\text{ atau }}(1,\infty )}$.

b. ${\textstyle {\frac {|x|+2}{x}}\leq 3}$

Syarat penyebut

$x\neq 0$

Syarat nilai mutlak

$|x|=\left\{{\begin{array}{r}x&{\text{jika }}x\geq 0\\-x&{\text{jika }}x<0\end{array}}\right.$

Pertidaksamaan

${\begin{aligned}{\frac {|x|+2}{x}}\leq 3\\{\frac {|x|+2}{x}}-3\leq 0\\{\frac {|x|+2-3x}{x}}\leq 0\end{aligned}}$

Kasus I Untuk ${\textstyle x\geq 0}$,

${\begin{aligned}{\frac {x+2-3x}{x}}\leq 0\\{\frac {2-2x}{x}}\leq 0\end{aligned}}$

Harga nol:

<-----+     +----->
  (-) | (+) | (-)
------●-----●------
      0     1

Dengan menerapkan syarat-syarat yang berlaku, ${\textstyle {\text{HP}}=\{x\geq 1\}}$.

Kasus II Untuk ${\textstyle x<0}$,

${\begin{aligned}{\frac {-x+2-3x}{x}}\leq 0\\{\frac {2-4x}{x}}\leq 0\end{aligned}}$

Harga nol:

<-----+     +----->
  (-) | (+) | (-)
------●-----●------
      0    1/2

Dengan menerapkan syarat-syarat yang berlaku, ${\textstyle {\text{HP}}=\{x<0\}}$.

Jadi, ${\textstyle {\text{HP}}}$ setelah digabung adalah ${\textstyle \{x<0{\text{ atau }}x\geq 1\}}$ atau dalam notasi interval ${\textstyle (-\infty ,0){\text{ atau }}[1,\infty )}$.

c. ${\textstyle {\frac {4{\sqrt {x}}}{x^{2}+3}}\leq {\frac {1}{\sqrt {x}}}}$

Syarat penyebut

${\begin{aligned}x^{2}+3&\neq 0\\x&\in \mathbb {R} \end{aligned}}$

${\begin{aligned}{\sqrt {x}}&\neq 0\\x&\neq 0\end{aligned}}$

Syarat akar

$x\geq 0$

Pertidaksamaan

${\begin{aligned}{\frac {4{\sqrt {x}}}{x^{2}+3}}&\leq {\frac {1}{\sqrt {x}}}\\{\frac {4{\sqrt {x}}}{x^{2}+3}}-{\frac {1}{\sqrt {x}}}&\leq 0\\{\frac {(4{\sqrt {x}})({\sqrt {x}})-(x^{2}+3)}{(x^{2}+3){\sqrt {x}}}}&\leq 0\\{\frac {4x-x^{2}-3}{(x^{2}+3){\sqrt {x}}}}&\leq 0\\{\frac {-(x-1)(x-3)}{(x^{2}+3){\sqrt {x}}}}&\leq 0\end{aligned}}$

Harga nol:

<-----+     +----->
  (-) | (+) | (-)
------●-----●------
      1     3

Dengan menerapkan syarat-syarat yang berlaku, ${\textstyle {\text{HP}}=\{0<x\leq 1{\text{ atau }}x\geq 3\}}$ atau dalam notasi interval ${\textstyle (0,1]{\text{ atau }}[3,\infty )}$.

2. Komposisi Fungsi

Fungsi berikut diketahui.

$f(x)={\frac {x}{x+1}}\qquad g(x)={\frac {x-1}{x-2}}$

a. ${\textstyle (f\circ g)(x)}$

${\begin{aligned}(f\circ g)(x)&=f(g(x))\\&=f\left({\frac {x-1}{x-2}}\right)\\&={\frac {\color {red}{\frac {x-1}{x-2}}}{\color {red}{\frac {x-1}{x-2}}\color {black}{+1}}}\\&={\frac {x-1}{x-1+x-2}}\\(f\circ g)(x)&={\frac {x-1}{2x-3}}\end{aligned}}$

Domain

Syarat penyebut

${\begin{aligned}2x-3&\neq 0\\2x&\neq 3\\x&\neq {\frac {3}{2}}\end{aligned}}$

Jadi, domain dari ${\textstyle (f\circ g)(x)}$ adalah ${\textstyle \{x\neq {\frac {3}{2}},x\in \mathbb {R} \}}$.

b. ${\textstyle (g\circ f)(x)}$

${\begin{aligned}(g\circ f)(x)&=g(f(x))\\&=g\left({\frac {x}{x+1}}\right)\\&={\frac {\color {red}{\frac {x}{x+1}}\color {black}{-1}}{\color {red}{\frac {x}{x+1}}\color {black}{-2}}}\\&={\frac {x-(x+1)}{x-2(x+1)}}\\&={\frac {-1}{-x-2}}\\(g\circ f)(x)&={\frac {1}{2-x}}\end{aligned}}$

Domain

Syarat penyebut

${\begin{aligned}2-x&\neq 0\\x&\neq 2\end{aligned}}$

Jadi, domain dari ${\textstyle (g\circ f)(x)}$ adalah ${\textstyle \{x\neq 2,x\in \mathbb {R} \}}$.

3. Limit Fungsi

Catatan: ${\textstyle \lim _{x\to a}f(x)}$ sama dengan $\lim _{x\to a}f(x)$

a. ${\textstyle \lim _{x\to 1}{\frac {x-1}{{\sqrt {x}}-1}}}$

${\begin{aligned}&\;\lim _{x\to 1}{\frac {x-1}{{\sqrt {x}}-1}}\\=&\;\lim _{x\to 1}{\frac {x-1}{{\sqrt {x}}-1}}\times {\frac {{\sqrt {x}}+1}{{\sqrt {x}}+1}}\\=&\;\lim _{x\to 1}{\frac {(x-1)({\sqrt {x}}+1)}{(x-1)}}\\=&\;\lim _{x\to 1}{\sqrt {x}}+1\\=&\;{\sqrt {1}}+1\\=&\;1+1\\=&\;2\end{aligned}}$

b. ${\textstyle \lim _{x\to 0}x^{2}\cos {\frac {1}{x}}}$

Karena ${\textstyle \lim _{x\to 0}\cos {\frac {1}{x}}}$ tidak ada, limit di atas tidak bisa ditentukan. Namun, dengan definisi dari fungsi kosinus, ${\textstyle -1\leq \cos \left({\frac {1}{x}}\right)\leq 1}$, kita bisa susun menjadi di bawah.

${\begin{array}{c}-1&\leq &\cos \left({\frac {1}{x}}\right)&\leq &1\\-x^{2}&\leq &x^{2}\cos \left({\frac {1}{x}}\right)&\leq &x^{2}\end{array}}$

Kita tahu bahwa ${\textstyle \lim _{x\to 0}-x^{2}=\lim _{x\to 0}x^{2}=0}$.

$0\leq x^{2}\cos \left({\frac {1}{x}}\right)\leq 0$

${\text{HP}}=[0,0]=\{0\}$

Kita dapat menyimpulkan bahwa ${\textstyle \lim _{x\to 0}x^{2}\cos {\frac {1}{x}}=0}$.

c. ${\textstyle \lim _{x\to 0}{\frac {1-\cos ^{2}(x)}{x^{2}}}}$

${\begin{aligned}&\;\lim _{x\to 0}{\frac {1-\cos ^{2}(x)}{x^{2}}}\\=&\;\lim _{x\to 0}{\frac {\sin ^{2}(x)}{x^{2}}}\\=&\;\left(\lim _{x\to 0}{\frac {\sin(x)}{x}}\right)^{2}\\=&\;(1)^{2}\\=&\;1\end{aligned}}$

4. Turunan Fungsi

a. ${\textstyle f(x)={\sqrt[{4}]{\frac {x^{2}+1}{x^{2}-1}}}}$

Catatan:

${\frac {d}{dx}}{\frac {U}{V}}={\frac {U'V-UV'}{V^{2}}}$

Turunan

${\begin{aligned}f'(x)&={\frac {d}{dx}}{\sqrt[{4}]{\frac {x^{2}+1}{x^{2}-1}}}\\f'(x)&={\frac {d}{dx}}\left({\frac {x^{2}+1}{x^{2}-1}}\right)^{\frac {1}{4}}\end{aligned}}$

${\begin{array}{l}U&=&x^{2}+1&&V&=&x^{2}-1\\U'&=&2x&&V'&=&2x\end{array}}$

${\begin{aligned}f'(x)&={\frac {d}{dx}}\left({\frac {U}{V}}\right)^{\frac {1}{4}}\\f'(x)&={\frac {1}{4}}\left({\frac {U}{V}}\right)^{-{\frac {3}{4}}}\left({\frac {U'V-UV'}{V^{2}}}\right)\\f'(x)&={\frac {U'V-UV'}{4V^{2}\left({\frac {U}{V}}\right)^{\frac {3}{4}}}}\\f'(x)&={\frac {2x(x^{2}-1)-2x(x^{2}+1)}{4(x^{2}-1)^{2}\left({\frac {x^{2}+1}{x^{2}-1}}\right)^{\frac {3}{4}}}}\\f'(x)&={\frac {2x(x^{2}-1-x^{2}-1)}{4(x^{2}-1)^{2}\left({\frac {x^{2}+1}{x^{2}-1}}\right)^{\frac {3}{4}}}}\\f'(x)&={\frac {-4x}{4(x^{2}-1)^{2}\left({\frac {x^{2}+1}{x^{2}-1}}\right)^{\frac {3}{4}}}}\\f'(x)&=-{\frac {x}{(x^{2}-1)^{2}\left({\frac {x^{2}+1}{x^{2}-1}}\right)^{\frac {3}{4}}}}\end{aligned}}$

b. ${\textstyle f(x)={\sqrt {x}}e^{x^{2}}(x^{2}+1)^{10}}$

$f'(x)={\frac {d}{dx}}{\sqrt {x}}e^{x^{2}}(x^{2}+1)^{10}$

Catatan:

${\frac {d}{dx}}e^{a}=e^{a}{\frac {da}{dx}}$

${\frac {d}{dx}}UV=U'V+UV'$

Turunan

${\begin{aligned}U&={\sqrt {x}}\\U'&={\frac {1}{2{\sqrt {x}}}}\\V&=e^{x^{2}}(x^{2}+1)^{10}\\V'&=2e^{x^{2}}x(x^{2}+1)^{10}+20e^{x^{2}}x(x^{2}+1)^{9}\\V'&=2e^{x^{2}}x(x^{2}+1)(x^{2}+1)^{9}+20e^{x^{2}}x(x^{2}+1)^{9}\\V'&=2e^{x^{2}}x(x^{2}+1)^{9}(x^{2}+1+10)\\V'&=2e^{x^{2}}x(x^{2}+1)^{9}(x^{2}+11)\end{aligned}}$

${\begin{aligned}f'(x)&=U'V+UV'\\f'(x)&={\frac {e^{x^{2}}(x^{2}+1)^{10}}{2{\sqrt {x}}}}+2e^{x^{2}}x{\sqrt {x}}(x^{2}+1)^{9}(x^{2}+11)\\f'(x)&={\frac {e^{x^{2}}(x^{2}+1)(x^{2}+1)^{9}+4e^{x^{2}}x^{2}(x^{2}+1)^{9}(x^{2}+11)}{2{\sqrt {x}}}}\\f'(x)&={\frac {e^{x^{2}}(x^{2}+1)^{9}(x^{2}+1+4x^{2}(x^{2}+11))}{2{\sqrt {x}}}}\\f'(x)&={\frac {e^{x^{2}}(x^{2}+1)^{9}(4x^{4}+45x^{2}+1)}{2{\sqrt {x}}}}\end{aligned}}$

5. Garis Singgung

$(x^{2}+y^{2}-1)^{3}=x^{2}y^{3}$ (grafik)

Catatan:

${\frac {d}{dx}}\;q={\frac {d}{dq}}\;q\;\;{\frac {dq}{dx}}$

${\frac {d}{dx}}UV=U'V+UV'$

Turunan terhadap $x$

${\frac {d}{dx}}(x^{2}+y^{2}-1)^{3}={\frac {d}{dx}}x^{2}y^{3}$

Misalkan

${\begin{aligned}u&=x^{2}+y^{2}-1\\{\frac {du}{dx}}&=2x+2y{\frac {dy}{dx}}\end{aligned}}$

Sehingga

${\begin{aligned}{\frac {d}{du}}\;u^{3}\;\;{\frac {du}{dx}}&=\left({\frac {d}{dx}}x^{2}\right)y^{3}+x^{2}\left({\frac {d}{dx}}y^{3}\right)\\3u^{2}(2x+2y{\frac {dy}{dx}})&=2xy^{3}+3x^{2}y^{2}{\frac {dy}{dx}}\\6u^{2}x+6u^{2}y{\frac {dy}{dx}}&=2xy^{3}+3x^{2}y^{2}{\frac {dy}{dx}}\\(6u^{2}y-3x^{2}y^{2}){\frac {dy}{dx}}&=2xy^{3}-6u^{2}x\\{\frac {dy}{dx}}&={\frac {2xy^{3}-6u^{2}x}{6u^{2}y-3x^{2}y^{2}}}\\{\frac {dy}{dx}}&={\frac {2xy^{3}-6x(x^{2}+y^{2}-1)^{2}}{6y(x^{2}+y^{2}-1)^{2}-3x^{2}y^{2}}}\\\end{aligned}}$

Perpotongan kurva dengan $x=1$

${\begin{aligned}(x^{2}+y^{2}-1)^{3}&=x^{2}y^{3}\\((1)^{2}+y^{2}-1)^{3}&=(1)^{2}y^{3}\\y^{6}&=y^{3}\\y^{6}-y^{3}&=0\\y^{3}(y^{3}-1)&=0\\y=0&\lor y=1\end{aligned}}$

Kita ambil titik ${\text{A}}(1,1)$.

${\begin{aligned}m&={\frac {dy}{dx}}\\m&={\frac {2xy^{3}-6x(x^{2}+y^{2}-1)^{2}}{6y(x^{2}+y^{2}-1)^{2}-3x^{2}y^{2}}}\\m&={\frac {2(1)(1)^{3}-6(1)((1)^{2}+(1)^{2}-1)^{2}}{6(1)((1)^{2}+(1)^{2}-1)^{2}-3(1)^{2}(1)^{2}}}\\m&={\frac {2-6}{6-3}}\\m&=-{\frac {4}{3}}\\\end{aligned}}$

Persamaan garis singgung

${\begin{aligned}y-y_{1}&=m(x-x_{1})\\y-1&=-{\frac {4}{3}}(x-1)\\y&=-{\frac {4}{3}}x+{\frac {4}{3}}+1\\y&=-{\frac {4}{3}}x+{\frac {7}{3}}\\4x+3y-7&=0\\\end{aligned}}$

(grafik)

6. Linearisasi Fungsi

${\begin{aligned}f(x)&={\sqrt {1-x}}\\f'(x)&=-{\frac {1}{2{\sqrt {1-x}}}}\end{aligned}}$

Fungsi linear ${\textstyle g(x)}$ adalah hasil linearisasi fungsi ${\textstyle f(x)}$ di sekitar ${\textstyle a}$.

$g(x)=f'(a)x+c$

Cari nilai ${\textstyle c}$ untuk ${\textstyle a=0}$

${\begin{aligned}g(0)&=f(0)\\f'(0)(0)+c&={\sqrt {1-0}}\\c&=1\end{aligned}}$

Fungsi linear ${\textstyle g(x)}$ di sekitar ${\textstyle a=0}$

${\begin{aligned}g(x)&=f'(0)x+1\\&=-{\frac {x}{2{\sqrt {1-0}}}}+1\\g(x)&=1-{\frac {x}{2}}\end{aligned}}$

Kasus I Pendekatan dari ${\textstyle {\sqrt {0,\!9}}}$

${\begin{aligned}{\sqrt {1-x}}&={\sqrt {0,\!9}}\\1-x&=0,\!9\\x&=0,\!1\end{aligned}}$

${\begin{aligned}g(0,\!1)&=1-{\frac {0,\!1}{2}}\\&=1-0,\!05\\g(0,\!1)&=0,\!95\\\end{aligned}}$

Jadi, pendekatan nilai dari ${\textstyle {\sqrt {0,\!9}}}$ adalah ${\textstyle 0,\!95}$.

Kasus II Pendekatan dari ${\textstyle {\sqrt {0,\!99}}}$

${\begin{aligned}{\sqrt {1-x}}&={\sqrt {0,\!99}}\\1-x&=0,\!99\\x&=0,\!01\end{aligned}}$

${\begin{aligned}g(0,\!01)&=1-{\frac {0,\!01}{2}}\\&=1-0,\!005\\g(0,\!01)&=0,\!995\\\end{aligned}}$

Jadi, pendekatan nilai dari ${\textstyle {\sqrt {0,\!99}}}$ adalah ${\textstyle 0,\!995}$.

7. Deret Taylor

${\begin{aligned}f(x)&={\frac {1}{\sqrt {x+1}}}\\f(x)&=(x+1)^{-{\frac {1}{2}}}\\f'(x)&=-{\frac {1}{2}}(x+1)^{-{\frac {3}{2}}}\\f''(x)&={\frac {3}{4}}(x+1)^{-{\frac {5}{2}}}\\f'''(x)&=-{\frac {15}{8}}(x+1)^{-{\frac {7}{2}}}\end{aligned}}$

Deret Taylor dari ${\textstyle f(x)}$ di sekitar ${\textstyle a=0}$:

${\begin{aligned}&\;f(a)+f'(a)(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+\dots \\=&\;f(0)+f'(0)(x-0)+{\frac {f''(0)}{2!}}(x-0)^{2}+{\frac {f'''(0)}{3!}}(x-0)^{3}+\dots \\=&\;f(0)+f'(0)x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f'''(0)}{3!}}x^{3}+\dots \\=&\;(0+1)^{-{\frac {1}{2}}}+\left(-{\frac {1}{2}}(0+1)^{-{\frac {3}{2}}}\right)x+{\frac {{\frac {3}{4}}(0+1)^{-{\frac {5}{2}}}}{2!}}x^{2}+{\frac {-{\frac {15}{8}}(0+1)^{-{\frac {7}{2}}}}{3!}}x^{3}+\dots \\=&\;1-{\frac {1}{2}}x+{\frac {\frac {3}{4}}{2!}}x^{2}-{\frac {\frac {15}{8}}{3!}}x^{3}+\dots \\=&\;1-{\frac {1}{2}}x+{\frac {3}{8}}x^{2}-{\frac {5}{16}}x^{3}+\dots \end{aligned}}$

8. Luas Maksimum Bangun Datar

       |
       |
       A
      /|\
     / | \
  B +--|--+ C
    |  |  |
    |  |  |
----+--0--+---- (sb. x)
       |
       |
    (sb. y)

Koordinat titik-titik segitiga sama kaki

${\text{A}}(0,12)\quad {\text{B}}(-p,12-p^{2})\quad {\text{C}}(p,12-p^{2})$

Syarat

$0<p<{\sqrt {12}}$

Luas bangun

${\begin{aligned}{\text{Luas segitiga}}&={\frac {1}{2}}\times 2p\times p^{2}=p^{3}\\{\text{Luas persegi panjang}}&=2p\times (12-p^{2})=24p-2p^{3}\\{\text{Total luas }}(L)&=-p^{3}+24p\end{aligned}}$

Titik puncak/belok

${\begin{aligned}L&=-p^{3}+24p\\{\frac {dL}{dp}}&=-3p^{2}+24\\0&=-3p^{2}+24\\3p^{2}&=24\\p&=2{\sqrt {2}}\\\end{aligned}}$

Luas maksimum

${\begin{aligned}(p=2{\sqrt {2}})\rightarrow L&=-p^{3}+24p\\L&=-(2{\sqrt {2}})^{3}+48{\sqrt {2}}\\L&=-16{\sqrt {2}}+48{\sqrt {12}}\\L&=32{\sqrt {2}}\\L&\approx 45,\!255\end{aligned}}$