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 setter inn x = 1 x = 1 x = 1
f ( 1 ) = 1 + 6 + 3 − 10 = 0 ✓ f(1) = 1 + 6 + 3 - 10 = 0 \checkmark f ( 1 ) = 1 + 6 + 3 − 10 = 0 ✓ Siden x = 1 x = 1 x = 1 ( x − 1 ) (x-1) ( x − 1 )
( x 3 + 6 x 2 + 3 x − 10 ) ÷ ( x − 1 ) = x 2 + 7 x + 10 (x^3 + 6x^2 + 3x - 10) \div (x-1) = x^2 + 7x + 10 ( x 3 + 6 x 2 + 3 x − 10 ) ÷ ( x − 1 ) = x 2 + 7 x + 10 Vi faktoriserer: x 2 + 7 x + 10 = ( x + 2 ) ( x + 5 ) x^2 + 7x + 10 = (x+2)(x+5) x 2 + 7 x + 10 = ( x + 2 ) ( x + 5 )
Nullpunktene er x = 1 , x = − 2 , x = − 5 ‾ ‾ \underline{\underline{x = 1, \; x = -2, \; x = -5}} x = 1 , x = − 2 , x = − 5
b)
Vi deriverer:
f ′ ( x ) = 3 x 2 + 12 x + 3 f'(x) = 3x^2 + 12x + 3 f ′ ( x ) = 3 x 2 + 12 x + 3 Vi setter f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0
3 x 2 + 12 x + 3 = 0 ⟹ x 2 + 4 x + 1 = 0 3x^2 + 12x + 3 = 0 \implies x^2 + 4x + 1 = 0 3 x 2 + 12 x + 3 = 0 ⟹ x 2 + 4 x + 1 = 0 x = − 4 ± 16 − 4 2 = − 4 ± 12 2 = − 2 ± 3 x = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3} x = 2 − 4 ± 16 − 4 = 2 − 4 ± 12 = − 2 ± 3 Vi sjekker med f ′ ′ ( x ) = 6 x + 12 f''(x) = 6x + 12 f ′′ ( x ) = 6 x + 12
f ′ ′ ( − 2 − 3 ) = 6 ( − 2 − 3 ) + 12 = − 6 3 < 0 f''(-2-\sqrt{3}) = 6(-2-\sqrt{3}) + 12 = -6\sqrt{3} < 0 f ′′ ( − 2 − 3 ) = 6 ( − 2 − 3 ) + 12 = − 6 3 < 0 toppunkt i x = − 2 − 3 ‾ ‾ \underline{\underline{x = -2 - \sqrt{3}}} x = − 2 − 3 f ′ ′ ( − 2 + 3 ) = 6 ( − 2 + 3 ) + 12 = 6 3 > 0 f''(-2+\sqrt{3}) = 6(-2+\sqrt{3}) + 12 = 6\sqrt{3} > 0 f ′′ ( − 2 + 3 ) = 6 ( − 2 + 3 ) + 12 = 6 3 > 0 bunnpunkt i x = − 2 + 3 ‾ ‾ \underline{\underline{x = -2 + \sqrt{3}}} x = − 2 + 3
c)
Vendepunktet ligger der f ′ ′ ( x ) = 0 f''(x) = 0 f ′′ ( x ) = 0
6 x + 12 = 0 ⟹ x = − 2 6x + 12 = 0 \implies x = -2 6 x + 12 = 0 ⟹ x = − 2 f ( − 2 ) = ( − 2 ) 3 + 6 ( − 2 ) 2 + 3 ( − 2 ) − 10 = − 8 + 24 − 6 − 10 = 0 f(-2) = (-2)^3 + 6(-2)^2 + 3(-2) - 10 = -8 + 24 - 6 - 10 = 0 f ( − 2 ) = ( − 2 ) 3 + 6 ( − 2 ) 2 + 3 ( − 2 ) − 10 = − 8 + 24 − 6 − 10 = 0 Stigningstallet til vendetangenten:
f ′ ( − 2 ) = 3 ⋅ 4 + 12 ⋅ ( − 2 ) + 3 = 12 − 24 + 3 = − 9 f'(-2) = 3 \cdot 4 + 12 \cdot (-2) + 3 = 12 - 24 + 3 = -9 f ′ ( − 2 ) = 3 ⋅ 4 + 12 ⋅ ( − 2 ) + 3 = 12 − 24 + 3 = − 9 Vendetangenten: y − 0 = − 9 ( x − ( − 2 ) ) y - 0 = -9(x - (-2)) y − 0 = − 9 ( x − ( − 2 ))
y = − 9 x − 18 ‾ ‾ \underline{\underline{y = -9x - 18}} y = − 9 x − 18
d)
Grafen har nullpunkter i x = − 5 x = -5 x = − 5 x = − 2 x = -2 x = − 2 x = 1 x = 1 x = 1 x = − 2 − 3 ≈ − 3,7 x = -2-\sqrt{3} \approx -3{,}7 x = − 2 − 3 ≈ − 3 , 7 x = − 2 + 3 ≈ − 0,3 x = -2+\sqrt{3} \approx -0{,}3 x = − 2 + 3 ≈ − 0 , 3 ( − 2 , 0 ) (-2, \, 0) ( − 2 , 0 ) y = − 9 x − 18 y = -9x - 18 y = − 9 x − 18 x x x f ( x ) → + ∞ f(x) \to +\infty f ( x ) → + ∞ x → − ∞ x \to -\infty x → − ∞ f ( x ) → − ∞ f(x) \to -\infty f ( x ) → − ∞