7.1 Bepaal f′(x) vanaf eerste beginsels indien f(x) = x² + x. 7.2 Bepaal f′(x) indien f(x) = 2x³ - 3x² + 8x. 7.3 Die raaklyn aan g(x) = ax³ + bx² + c het 'n minimum helling (gradiënt) by die punt (−1 ; −7). Vir watter waardes van x sal g konkaf op wees?

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7.1 Bepaal f′(x) vanaf eerste beginsels indien f(x) = x² + x - NSC Mathematics - Question 7 - 2022 - Paper 1

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7.1 Bepaal f′(x) vanaf eerste beginsels indien f(x) = x² + x. 7.2 Bepaal f′(x) indien f(x) = 2x³ - 3x² + 8x. 7.3 Die raaklyn aan g(x) = ax³ + bx² + c het 'n minimu... show full transcript

Worked Solution & Example Answer:7.1 Bepaal f′(x) vanaf eerste beginsels indien f(x) = x² + x - NSC Mathematics - Question 7 - 2022 - Paper 1

Step 1

Bepaal f′(x) vanaf eerste beginsels indien f(x) = x² + x.

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Answer

To find the derivative f′(x) from first principles, we use the definition of a derivative:

$f′(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Substituting the function: $f′(x) = \lim_{h \to 0} \frac{((x + h)² + (x + h)) - (x² + x)}{h}$

Expanding the numerator: $= \lim_{h \to 0} \frac{(x² + 2xh + h² + x + h) - (x² + x)}{h}$ $= \lim_{h \to 0} \frac{2xh + h² + h}{h}$

Factoring out h: $= \lim_{h \to 0} (2x + h + 1)$

As h approaches 0, we have: $f′(x) = 2x + 1$

Step 2

Bepaal f′(x) indien f(x) = 2x³ - 3x² + 8x.

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Answer

Using the power rule for differentiation: $f′(x) = 3(2x²) - 2(3x) + 8$

Calculating gives: $f′(x) = 6x² - 6x + 8$

Step 3

Die raaklyn aan g(x) = ax³ + bx² + c het 'n minimum helling (gradiënt) by die punt (−1 ; −7).

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Answer

To find where the graph is concave up, we need to analyze the second derivative.

First, we find the first derivative: $g′(x) = 3ax² + 2bx$

Setting the first derivative to zero at the point (−1) for a minimum: $g′(−1) = 3a(-1)² + 2b(-1) = 0 \Rightarrow 3a - 2b = 0$

This simplifies to: $b = \frac{3}{2}a$

Next, for concavity: We need the second derivative: $g′′(x) = 6ax + 2b$

For concave up, set: $g′′(−1) > 0 \Rightarrow 6a(-1) + 2b > 0$

Substituting b: $-6a + 3a > 0 \Rightarrow -3a > 0 \Rightarrow a < 0$

Therefore, for values of x where g is concave up, we conclude: $x > -1$

7.1 Bepaal f′(x) vanaf eerste beginsels indien f(x) = x² + x - NSC Mathematics - Question 7 - 2022 - Paper 1

Worked Solution & Example Answer:7.1 Bepaal f′(x) vanaf eerste beginsels indien f(x) = x² + x - NSC Mathematics - Question 7 - 2022 - Paper 1

Bepaal f′(x) vanaf eerste beginsels indien f(x) = x² + x.

Bepaal f′(x) indien f(x) = 2x³ - 3x² + 8x.

Die raaklyn aan g(x) = ax³ + bx² + c het 'n minimum helling (gradiënt) by die punt (−1 ; −7).

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