SNBT Zone : Maksimum dan Minimum
Table of Contents
Tipe:
No.
Nilai maksimum dari- 0
- 1
- 2
- 3
- 4
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
f(x)&={^2\negmedspace\log}(x-1)+{^2\negmedspace\log}(-x-3)\\
&={^2\negmedspace\log}\left((x-1)(-x-3)\right)\\
&={^2\negmedspace\log}\left(-x^2-3x+x+3\right)\\
&={^2\negmedspace\log}\left(-x^2-2x+3\right)
\end{aligned}\)
Karena fungsi log dengan basis lebih dari 1 adalah fungsi naik, maka kita cari dahulu nilai maksimum dari−x2 − 2x + 3
Misalp = −x2 − 2x + 3
mencari pmax berartip' = 0
\(\begin{aligned} -2x-2&=0\\ -2x&=2\\ x&=-1 \end{aligned}\)
\(\begin{aligned} f(x)_{\max}&={^2\negmedspace\log}\left(-(-1)^2-2(-1)+3\right)\\ &={^2\negmedspace\log}\left(-1+2+3\right)\\ &={^2\negmedspace\log}4\\ &=\boxed{\boxed{2}} \end{aligned}\)
Karena fungsi log dengan basis lebih dari 1 adalah fungsi naik, maka kita cari dahulu nilai maksimum dari
Misal
mencari pmax berarti
\(\begin{aligned} -2x-2&=0\\ -2x&=2\\ x&=-1 \end{aligned}\)
\(\begin{aligned} f(x)_{\max}&={^2\negmedspace\log}\left(-(-1)^2-2(-1)+3\right)\\ &={^2\negmedspace\log}\left(-1+2+3\right)\\ &={^2\negmedspace\log}4\\ &=\boxed{\boxed{2}} \end{aligned}\)
Jadi, nilai maksimum dari f(x) = 2log(x − 1) + 2log(−x − 3) adalah 2.
JAWAB: C
JAWAB: C
No.
Jika- 6
- 5
- 4
- 3
- 2
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\left(f\circ g\right)(x)&=f\left(g(x)\right)\\
&=f\left(-2x+4\right)\\
&=3-(-2x+4)^2
\end{aligned}\)
\(\begin{aligned} \left(f\circ g\right)'(x)&=0\\ -2(-2x+4)(-2)&=0\\ -8x+16&=0\\ 8x&=16\\ x&=2 \end{aligned}\)
\(\begin{aligned} \left(f\circ g\right)(2)&=3-(-2(2)+4)^2\\ &=3-(-4+4)^2\\ &=3-(0)^2\\ =\boxed{\boxed{3}} \end{aligned}\)
\(\begin{aligned} \left(f\circ g\right)'(x)&=0\\ -2(-2x+4)(-2)&=0\\ -8x+16&=0\\ 8x&=16\\ x&=2 \end{aligned}\)
\(\begin{aligned} \left(f\circ g\right)(2)&=3-(-2(2)+4)^2\\ &=3-(-4+4)^2\\ &=3-(0)^2\\ =\boxed{\boxed{3}} \end{aligned}\)
Jadi, nilai minimum fungsi (f∘g)(x) adalah 3.
JAWAB: D
JAWAB: D
No.
Jika- −3
- −4
- 5
- 6
- 9
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
4a&=-4\\
a&=-1
\end{aligned}\)
\(\begin{aligned} f(x)&=-x^3+x^2+x-4\\ f'(x)&=-3x^2+2x+1 \end{aligned}\)
fmax ⟶ f' = 0
\(\begin{aligned} -3x^2+2x+1&=0\\ 3x^2-2x-1&=0\\ (3x+1)(x-1)&=0 \end{aligned}\)
x=-\dfrac13 (TM) dan x = 1 .
\(\begin{aligned} f(1)&=-1^3+1^2+1-4\\ &=\boxed{\boxed{-3}} \end{aligned}\)
\(\begin{aligned} f(x)&=-x^3+x^2+x-4\\ f'(x)&=-3x^2+2x+1 \end{aligned}\)
\(\begin{aligned} -3x^2+2x+1&=0\\ 3x^2-2x-1&=0\\ (3x+1)(x-1)&=0 \end{aligned}\)
\(\begin{aligned} f(1)&=-1^3+1^2+1-4\\ &=\boxed{\boxed{-3}} \end{aligned}\)
Jadi, nilai
maksimum f(x) untuk 0 ≤ x ≤ 2 adalah −3.
JAWAB: A
JAWAB: A
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