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Parents Pricing Home NSC Mathematics Finance growth and decay 8.1 Gegee $f(x)=3-2x^2$: Bepaal $f'(x)$, vanuit eerste beginsels
8.1 Gegee $f(x)=3-2x^2$: Bepaal $f'(x)$, vanuit eerste beginsels - NSC Mathematics - Question 8 - 2017 - Paper 1 Question 8
View full question 8.1 Gegee $f(x)=3-2x^2$: Bepaal $f'(x)$, vanuit eerste beginsels.
8.2 Bepaal $\frac{dy}{dx}$ as $y=\frac{12x^2+2x+1}{6x}$.
8.3 Die funksie $f(x)=x^3+bx^2+cx-4$ het... show full transcript
View marking scheme Worked Solution & Example Answer:8.1 Gegee $f(x)=3-2x^2$: Bepaal $f'(x)$, vanuit eerste beginsels - NSC Mathematics - Question 8 - 2017 - Paper 1
Bepaal $f'(x)$, vanuit eerste beginsels. Only available for registered users.
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To find the derivative f ′ ( x ) f'(x) f ′ ( x )
f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} f ′ ( x ) = lim h → 0 h f ( x + h ) − f ( x )
Compute f ( x + h ) f(x+h) f ( x + h ) f ( x + h ) = 3 − 2 ( x + h ) 2 = 3 − 2 ( x 2 + 2 x h + h 2 ) = 3 − 2 x 2 − 4 x h − 2 h 2 f(x+h) = 3 - 2(x+h)^2 = 3 - 2(x^2 + 2xh + h^2) = 3 - 2x^2 - 4xh - 2h^2 f ( x + h ) = 3 − 2 ( x + h ) 2 = 3 − 2 ( x 2 + 2 x h + h 2 ) = 3 − 2 x 2 − 4 x h − 2 h 2
Substitute in the limit:
f ′ ( x ) = lim h → 0 ( 3 − 2 x 2 − 4 x h − 2 h 2 ) − ( 3 − 2 x 2 ) h f'(x) = \lim_{h \to 0} \frac{(3 - 2x^2 - 4xh - 2h^2) - (3 - 2x^2)}{h} f ′ ( x ) = lim h → 0 h ( 3 − 2 x 2 − 4 x h − 2 h 2 ) − ( 3 − 2 x 2 )
Simplify:
f ′ ( x ) = lim h → 0 − 4 x h − 2 h 2 h = lim h → 0 ( − 4 x − 2 h ) = − 4 x f'(x) = \lim_{h \to 0} \frac{-4xh - 2h^2}{h} = \lim_{h \to 0} (-4x - 2h) = -4x f ′ ( x ) = lim h → 0 h − 4 x h − 2 h 2 = lim h → 0 ( − 4 x − 2 h ) = − 4 x
Thus, f ′ ( x ) = − 4 x f'(x) = -4x f ′ ( x ) = − 4 x
Bepaal $\frac{dy}{dx}$ as $y=\frac{12x^2+2x+1}{6x}$. Only available for registered users.
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To differentiate the function, we apply the quotient rule:
y = 12 x 2 + 2 x + 1 6 x y = \frac{12x^2 + 2x + 1}{6x} y = 6 x 12 x 2 + 2 x + 1
First, simplify the expression:
y = 12 x 2 6 x + 2 x 6 x + 1 6 x = 2 x + 1 3 + 1 6 x y = \frac{12x^2}{6x} + \frac{2x}{6x} + \frac{1}{6x} = 2x + \frac{1}{3} + \frac{1}{6x} y = 6 x 12 x 2 + 6 x 2 x + 6 x 1 = 2 x + 3 1 + 6 x 1
Now differentiate:
d y d x = 2 − 1 6 x 2 \frac{dy}{dx} = 2 - \frac{1}{6x^2} d x d y = 2 − 6 x 2 1
Therefore, d y d x = 2 − 1 6 x 2 \frac{dy}{dx} = 2 - \frac{1}{6x^2} d x d y = 2 − 6 x 2 1
Bereken die waardes van $b$ en $c$. Only available for registered users.
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Given the function f ( x ) = x 3 + b x 2 + c x − 4 f(x) = x^3 + bx^2 + cx - 4 f ( x ) = x 3 + b x 2 + c x − 4 b b b c c c ( 2 ; 4 ) (2; 4) ( 2 ; 4 )
Calculate the first derivative:
f ′ ( x ) = 3 x 2 + 2 b x + c f'(x) = 3x^2 + 2bx + c f ′ ( x ) = 3 x 2 + 2 b x + c f ′ ( 2 ) = 0 f'(2) = 0 f ′ ( 2 ) = 0 3 ( 2 ) 2 + 2 b ( 2 ) + c = 0 3(2)^2 + 2b(2) + c = 0 3 ( 2 ) 2 + 2 b ( 2 ) + c = 0 12 + 4 b + c = 0 12 + 4b + c = 0 12 + 4 b + c = 0
Calculate f ( 2 ) f(2) f ( 2 ) f ( 2 ) = ( 2 ) 3 + b ( 2 ) 2 + c ( 2 ) − 4 = 4 f(2) = (2)^3 + b(2)^2 + c(2) - 4 = 4 f ( 2 ) = ( 2 ) 3 + b ( 2 ) 2 + c ( 2 ) − 4 = 4 8 + 4 b + 2 c − 4 = 4 8 + 4b + 2c - 4 = 4 8 + 4 b + 2 c − 4 = 4 4 + 4 b + 2 c = 4 4 + 4b + 2c = 4 4 + 4 b + 2 c = 4 4 b + 2 c = 0 4b + 2c = 0 4 b + 2 c = 0
Solve the system of equations:
From Equation 2: 2 b + c = 0 2b + c = 0 2 b + c = 0 c = − 2 b c = -2b c = − 2 b
ightarrow 12 + 2b = 0$$
ightarrow b = -6$$
Substitute $b$ back to find $c$:
$$c = -2(-6) = 12$$
Thus, the values are $b = -6$ and $c = 12$. Join the NSC students using SimpleStudy...97% of StudentsReport Improved Results
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