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SMK Teknologi dan Rekayasa

Soal 119 Nilai Maksimum-Minimum Fungsi Aljabar


Nilai\ minimum\ fungsi\ f(x)=x^{2}-x\ pada\ interval\ -1\leq x\leq 3\ adalah\ ....
a.\ \ 1\ \ \ \ \ \ \ \ \ \ b.\ \ \frac{1}{2}\ \ \ \ \ \ \ \ \ \ c.\ \ 0\ \ \ \ \ \ \ \ \ \ d.\ -\frac{1}{2}\ \ \ \ \ \ \ \ \ \ e.\ -\frac{1}{4}
Jawaban: e

Nilai\ maksimum\ (atau\ minimum)\ fungsi\ dalam\ interval\ tertutup\ terjadi\\\mathbf{pada\ stasioner\ atau\ pada\ batas\ interval}.
f(x)=x^{2}-x\\f{}'(x)=2x-1

Stasioner fungsi:
\begin{array}{rcl}f{}'(x)&=&0\\2x-1&=&0\\2x&=&1\\x&=&\frac{1}{2}\end{array}

Nilai\ ekstrim\ fungsi:
f\big(\frac{1}{2}\big)=\big(\frac{1}{2}\big)^{2}-\frac{1}{2}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}
f(-1)=(-1)^{2}-(-1)=1+1=2
f(3)=(3)^{2}-3=9-3=6
Nilai\ minimumnya\ adalah\ -\frac{1}{4}.

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