KI-generert løsningsforslag. Dette løsningsforslaget er skrevet av en språkmodell (Claude) og er ikke verifisert av lærer. Det kan inneholde regnefeil, upresis bruk av begreper eller mangler.
a)
Vi skal finne nullpunktene til g ( x ) = 1 2 e x ⋅ ( 2 x − 1 ) 2 g(x) = \frac{1}{2}e^x \cdot (2x-1)^2 g ( x ) = 2 1 e x ⋅ ( 2 x − 1 ) 2
g ( x ) = 0 ⟺ 1 2 e x ⋅ ( 2 x − 1 ) 2 = 0 g(x) = 0 \iff \frac{1}{2}e^x \cdot (2x-1)^2 = 0 g ( x ) = 0 ⟺ 2 1 e x ⋅ ( 2 x − 1 ) 2 = 0 Siden 1 2 e x > 0 \frac{1}{2}e^x > 0 2 1 e x > 0 x x x ( 2 x − 1 ) 2 = 0 (2x-1)^2 = 0 ( 2 x − 1 ) 2 = 0
2 x − 1 = 0 ⟺ x = 1 2 2x - 1 = 0 \iff x = \frac{1}{2} 2 x − 1 = 0 ⟺ x = 2 1 g g g x = 1 2 ‾ ‾ \underline{\underline{x = \frac{1}{2}}} x = 2 1
b)
Vi bruker produktregelen på g ( x ) = u ( x ) ⋅ v ( x ) g(x) = u(x) \cdot v(x) g ( x ) = u ( x ) ⋅ v ( x )
u ( x ) = 1 2 e x , v ( x ) = ( 2 x − 1 ) 2 u(x) = \frac{1}{2}e^x, \qquad v(x) = (2x-1)^2 u ( x ) = 2 1 e x , v ( x ) = ( 2 x − 1 ) 2 u ′ ( x ) = 1 2 e x , v ′ ( x ) = 2 ( 2 x − 1 ) ⋅ 2 = 4 ( 2 x − 1 ) u'(x) = \frac{1}{2}e^x, \qquad v'(x) = 2(2x-1) \cdot 2 = 4(2x-1) u ′ ( x ) = 2 1 e x , v ′ ( x ) = 2 ( 2 x − 1 ) ⋅ 2 = 4 ( 2 x − 1 ) Produktregelen gir
g ′ ( x ) = u ′ v + u v ′ = 1 2 e x ( 2 x − 1 ) 2 + 1 2 e x ⋅ 4 ( 2 x − 1 ) g'(x) = u'v + uv' = \frac{1}{2}e^x(2x-1)^2 + \frac{1}{2}e^x \cdot 4(2x-1) g ′ ( x ) = u ′ v + u v ′ = 2 1 e x ( 2 x − 1 ) 2 + 2 1 e x ⋅ 4 ( 2 x − 1 ) Vi faktoriserer ut 1 2 e x ( 2 x − 1 ) \frac{1}{2}e^x(2x-1) 2 1 e x ( 2 x − 1 )
g ′ ( x ) = 1 2 e x ( 2 x − 1 ) [ ( 2 x − 1 ) + 4 ] = 1 2 e x ( 2 x − 1 ) ( 2 x + 3 ) g'(x) = \frac{1}{2}e^x(2x-1)\bigl[(2x-1) + 4\bigr] = \frac{1}{2}e^x(2x-1)(2x+3) g ′ ( x ) = 2 1 e x ( 2 x − 1 ) [ ( 2 x − 1 ) + 4 ] = 2 1 e x ( 2 x − 1 ) ( 2 x + 3 ) Dette er det vi skulle vise. □ \square □
c)
Vi setter g ′ ( x ) = 0 g'(x) = 0 g ′ ( x ) = 0 1 2 e x > 0 \frac{1}{2}e^x > 0 2 1 e x > 0 x x x
( 2 x − 1 ) ( 2 x + 3 ) = 0 (2x-1)(2x+3) = 0 ( 2 x − 1 ) ( 2 x + 3 ) = 0 x = 1 2 eller x = − 3 2 x = \frac{1}{2} \quad \text{eller} \quad x = -\frac{3}{2} x = 2 1 eller x = − 2 3 Vi bestemmer fortegnet til g ′ ( x ) = 1 2 e x ⋅ ( 2 x − 1 ) ⋅ ( 2 x + 3 ) g'(x) = \frac{1}{2}e^x \cdot \textcolor{steelblue}{(2x-1)} \cdot \textcolor{seagreen}{(2x+3)} g ′ ( x ) = 2 1 e x ⋅ ( 2 x − 1 ) ⋅ ( 2 x + 3 )
x < − 3 2 x < -\frac{3}{2} x < − 2 3 x = − 3 2 x = -\frac{3}{2} x = − 2 3 − 3 2 < x < 1 2 -\frac{3}{2} < x < \frac{1}{2} − 2 3 < x < 2 1 x = 1 2 x = \frac{1}{2} x = 2 1 x > 1 2 x > \frac{1}{2} x > 2 1 2 x − 1 \textcolor{steelblue}{2x-1} 2 x − 1 − - − − - − − - − 0 0 0 + + + 2 x + 3 \textcolor{seagreen}{2x+3} 2 x + 3 − - − 0 0 0 + + + + + + + + + g ′ ( x ) g'(x) g ′ ( x ) + + + 0 0 0 − - − 0 0 0 + + + g g g voksende topp avtagende bunn voksende
g g g toppunkt i x = − 3 2 x = -\frac{3}{2} x = − 2 3 bunnpunkt i x = 1 2 x = \frac{1}{2} x = 2 1
Vi beregner y y y
g ( 1 2 ) = 1 2 e 1 / 2 ⋅ ( 2 ⋅ 1 2 − 1 ) 2 = 1 2 e 1 / 2 ⋅ 0 = 0 g\!\left(\frac{1}{2}\right) = \frac{1}{2}e^{1/2}\cdot\left(2\cdot\frac{1}{2}-1\right)^2 = \frac{1}{2}e^{1/2}\cdot 0 = 0 g ( 2 1 ) = 2 1 e 1/2 ⋅ ( 2 ⋅ 2 1 − 1 ) 2 = 2 1 e 1/2 ⋅ 0 = 0 g ( − 3 2 ) = 1 2 e − 3 / 2 ⋅ ( 2 ⋅ ( − 3 2 ) − 1 ) 2 = 1 2 e − 3 / 2 ⋅ ( − 4 ) 2 = 1 2 e − 3 / 2 ⋅ 16 = 8 e − 3 / 2 g\!\left(-\frac{3}{2}\right) = \frac{1}{2}e^{-3/2}\cdot\left(2\cdot\left(-\frac{3}{2}\right)-1\right)^2 = \frac{1}{2}e^{-3/2}\cdot(-4)^2 = \frac{1}{2}e^{-3/2}\cdot 16 = 8e^{-3/2} g ( − 2 3 ) = 2 1 e − 3/2 ⋅ ( 2 ⋅ ( − 2 3 ) − 1 ) 2 = 2 1 e − 3/2 ⋅ ( − 4 ) 2 = 2 1 e − 3/2 ⋅ 16 = 8 e − 3/2 Koordinater:
Toppunkt: ( − 3 2 , 8 e − 3 / 2 ) ‾ ‾ \underline{\underline{\left(-\frac{3}{2},\ 8e^{-3/2}\right)}} ( − 2 3 , 8 e − 3/2 )
Bunnpunkt: ( 1 2 , 0 ) ‾ ‾ \underline{\underline{\left(\frac{1}{2},\ 0\right)}} ( 2 1 , 0 )