KUIS Diketahui fungsi [tex]\displaystyle \rm f(x) = a\sqrt{x}+\frac{b}{\sqrt{x}}[/tex] memiliki titi.... Question from @xcvi

KenJhenar

aplikasi turunan

f(x) = a√x + b/√x

stasioner → f'(x) = 0

f'(x) = 0

a/(2√x) - b/(2x√x) = 0

a - b/x = 0

(4,13)

a - b/4 = 0

4a - b = 0

b = 4a ... (1)

f(4) = 13

a√4 + b/√4 = 13

2a + 4a/2 = 13

4a = 13

a = 13/4

b = 4a = 13

f(x) = c√x + d/√x

f(4) = 13

2c + d/2 = 13 ... (1)

belok → f" = 0

f'(x) = c/(2√x) - d/(2x√x)

f"(x) = 0

-c/(4x√x) + 3d/(4x²√x) = 0

-c + 3d/x = 0

c = 3d/4 ... (2)

2c + d/2 = 13

2(3d/4) + d/2 = 13

2d = 13

d = 13/2

c = 3(13/2)/4 = 39/8

a + b - c - d

= 13/4 + 13 - 39/8 - 13/2

= 13(1/4 + 1 - 3/8 - 1/2)

= 13(2 + 8 - 3 - 4)/8

= 13 × 3/8

= 39/8

= 4 7/8

9 votes Thanks 20

xcvi benarr

suciislamiati498 keren

henriyulianto

Terbukti benar bahwa a + b - c - d = 4⅞.
 

Pembahasan

$\displaystyle f(x) = a\sqrt{x}+\frac{b}{\sqrt{x}}$ dan $\displaystyle g(x)=c\sqrt{4}+\frac{d}{\sqrt{4}}$ berturut-turut memiliki titik stasioner dan titik belok di (4, 13). Fungsi kedua diganti namanya menjadi $g(x)$ agar lebih jelas.

Maka:

$\begin{aligned}13&=f(4)&&=g(4)\\13&=a\sqrt{4}+\frac{b}{\sqrt{4}}&&=c\sqrt{4}+\frac{d}{\sqrt{4}}\\&=2a+\frac{b}{2}&&=2c+\frac{d}{2}\\\therefore\ 26&=\begin{cases}4a+b&(i)\\4c+d&(ii)\end{cases}\!\!\!\!\!\!\!\!\!\end{aligned}$

Karena (4, 13) adalah titik stasioner dari f(x), apapun jenisnya, maka $f'(4)=0$ .

$\begin{aligned}0&=\left.\left(a\sqrt{x}+\frac{b}{\sqrt{x}}\right)'\right|_{x=4}\\&=\left.\left(ax^{1/2}+bx^{-1/2}\right)'\right|_{x=4}\\&=\left.\left(\frac{ax^{-1/2}}{2}-\frac{bx^{-3/2}}{2}\right)\right|_{x=4}\\&=\left.\left(\frac{a}{2\sqrt{x}}-\frac{b}{2\sqrt{x^3}}\right)\right|_{x=4}\\&=\frac{a}{2\sqrt{4}}-\frac{b}{2\sqrt{64}}\\&=\frac{a}{4}-\frac{b}{16}\\0&=4a-b\ \Rightarrow\ 4a=b\quad(iii)\end{aligned}$

Karena (4, 13) adalah titik belok dari $g(x)$ , maka $g''(4) = 0$ . Bentuk fungsi $g(x)$ serupa dengan $f(x)$ , sehingga kita dapat memanfaatkan hasil di atas.

$\begin{aligned}0&=\left.\left(c\sqrt{x}+\frac{d}{\sqrt{x}}\right)''\right|_{x=4}\\&=\left.\left(\frac{cx^{-1/2}}{2}-\frac{dx^{-3/2}}{2}\right)'\right|_{x=4}\\&=\left.\left(-\frac{cx^{-3/2}}{4}+\frac{3dx^{-5/2}}{4}\right)\right|_{x=4}\\&=\left.\left(\frac{3d}{4\sqrt{x^5}}-\frac{c}{4\sqrt{x^3}}\right)\right|_{x=4}\\&=\frac{3d}{4\sqrt{4^5}}-\frac{c}{4\sqrt{4^3}}\\&=\frac{3d}{4\sqrt{2^{10}}}-\frac{c}{4\sqrt{2^{6}}}\end{aligned}$
$\begin{aligned}0&=\frac{3d}{4\cdot2^5}-\frac{c}{4\cdot2^3}\\0&=\frac{3d}{2^5}-\frac{c}{2^3}=\frac{3d}{32}-\frac{c}{8}\\0&=3d-4c\ \Rightarrow\ 4c=3d\quad(iv)\end{aligned}$

Substitusi $(iii) \to (i)$ memberikan

$\begin{aligned}&26=4a+b=b+b=2b\\&\Rightarrow b=13\end{aligned}$

Substitusi $(iv) \to (ii)$ memberikan

$\begin{aligned}&26=4c+d=3d+d=4d\\&\Rightarrow 4d=26\end{aligned}$

Kemudian, kita akan membuktikan bahwa a + b - c - d = 4⅞.

$\begin{aligned}&a + b - c - d\\{=\ }&a+b-(c+d)\\{=\ }&\frac{1}{4}\left[4a+4b-(4c+4d)\right]\\&...\textsf{ $b\to4a$ dan $3d\to4c$}\\{=\ }&\frac{1}{4}\left[b+4b-(3d+4d)\right]\\{=\ }&\frac{1}{4}\left[5b-\frac{7}{4}\cdot4d\right]\end{aligned}$
$\begin{aligned}&...\textsf{ $13\to b$ dan $26\to4d$}\\{=\ }&\frac{1}{4}\left[5\cdot13-\frac{7}{4}\cdot26\right]\\{=\ }&\frac{1}{4}\left[5\cdot13-\frac{7}{2}\cdot13\right]\\{=\ }&\frac{1}{8}\left[10\cdot13-7\cdot13\right]\\{=\ }&\frac{13}{8}\left[10-7\right]\\{=\ }&\frac{39}{8}=\frac{4\cdot8+7}{8}\\{=\ }&\bf4\,\frac{7}{8}\implies\sf terbukti!\end{aligned}$

$\blacksquare$

17 votes Thanks 26

xcvi benul

suciislamiati498 hebat jawaban nya lengkap dan membantu sekali

henriyulianto terima kasih kak @xcvi

henriyulianto terima kasih kak @suciislamiati498. masih kurang lengkap sih, namun selengkapnya bisa mencari di buku masing2 atau di web. :D

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