8.1 Determine f'(x) from first principles if f(x) = -x^2 + 4 - NSC Mathematics - Question 8 - 2016 - Paper 1

For y = 3x^3 + 10x:

To find the derivative, we apply the power rule: $\frac{dy}{dx} = \frac{d}{dx}(3x^3) + \frac{d}{dx}(10x)$ $= 9x^2 + 10$

To differentiate f(x) = (x - 3)^2/x, we can rewrite it as: f(x) = (x - 3)^2 \cdot x^{-1}. Using the product rule: $f'(x) = (2(x - 3) \cdot 1) \cdot x^{-1} + (x - 3)^2 \cdot (-x^{-2})$ $= \frac{2(x - 3)}{x} - \frac{(x - 3)^2}{x^2}$ Simplifying gives: $$f'(x) = \frac{2(x - 3) - (x - 3)^2}{x^2} = \frac{-x^2 + 9x - 6}{x^2}.$

To prove (x - 2) is a factor, we substitute x = 2 into f:

$f(2) = 2(2)^3 - 23(2)^2 + 80(2) - 84 = 0$ This means (x - 2) is a factor since f(2) = 0.

Since (x - 2) is a factor, we divide f(x) by (x - 2): $f(x) = 2x^3 - 23x^2 + 80x - 84$ Using synthetic or polynomial division, we find: $f(x) = (x - 2)(2x^2 - 19x + 42)$ Next, factor 2x^2 - 19x + 42 to get: $f(x) = (x - 2)(2x - 6)(x - 7)$

To find the turning points, we first find f'(x): f'(x) = 0 means we solve: $2(x - 2)(x - 7) = 0$ The x-coordinates of the turning points are x = 2 and x = 7.

The turning points are x = 2 and x = 7. Their corresponding y-values can be found by substituting back into f: At x = 2, f(2) = 0; At x = 7, f(7) = -25. Intercepts:

• y-intercept when x = 0 gives f(0) = -84. Sketch the graph with these points, showing the curve's behavior.

To find the point where the tangent has a slope of 40, we set f'(x) = 40. Solving: $6x^2 - 46x + 40 = 0$ Use the quadratic formula or factorization: This yields x = 1 and x = 20/3. The corresponding y-value can be determined using f: Substituting x = 1 gives f(1) = -25. Now, the equation of the tangent line at (1, -25) is: y - (-25) = 40(x - 1) Thus, to find the y-intercept (when x = 0):