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SMUP 2010

Solusi SMUP 2010 Kode 041 – Nomor 23


Supaya\ fungsi
f(x)=a\sqrt{x}+\frac{b}{\sqrt{x}}
mempunyai\ titik\ (4,\ 13)\ sebagai\ titik\ belok,\ maka\ nilai\ 8a+4b\\adalah\ ....
A.\quad  63\\B.\quad 65\\C.\quad 73\\D.\quad 85\\E.\quad 53
Jawaban: B

Syarat\ titik\ belok\quad \rightarrow \quad f''(x)=0
f(x)=a\sqrt{x}+\frac{b}{\sqrt{x}}=ax^{\frac{1}{2}}+bx^{-\frac{1}{2}}
f'(x)=\frac{1}{2}ax^{-\frac{1}{2}}-\frac{1}{2}bx^{-1\frac{1}{2}}
f''(x)=-\frac{1}{4}ax^{-1\frac{1}{2}}+\frac{3}{4}bx^{-2\frac{1}{2}}=-\frac{a}{4x\sqrt{x}}+\frac{3b}{4x^2\sqrt{x}}
Absis\ titik\ belok\ adalah\ x=4,\ maka:
\begin{array}{rcl}-\frac{a}{4x\sqrt{x}}+\frac{3b}{4x^2\sqrt{x}}&=&0\\\\\frac{a}{4x\sqrt{x}}&=&\frac{3b}{4x^2\sqrt{x}}\\\\a&=&\frac{3b}{x}\\\\a&=&\frac{3b}{4}\\\\4a&=&3b\quad .................\quad (*)\end{array}
Melalui\ titik\ (4,\ 13)\ diperoleh:
\begin{array}{rcl}f(4)&=&13\\\\2a+\frac{b}{2}&=&13\\\\4a+b&=&26\\\\3b+b&=&26\\\\4b&=&26\\\\b&=&6,5\end{array}
Jadi:

\begin{array}{rcl}8a+4b&=&2.4a+4b\\&=&2.3b+4b\\&=&10b\\&=&65\end{array}

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