Given $f(x) = 3 - 2x^2$ - NSC Mathematics - Question 8 - 2017 - Paper 1

To find rac{dy}{dx}, we apply the quotient rule since we have a quotient of functions:

If $y = \frac{u}{v}$, then:

$\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}$

Given:

• $u = 12x^2 + 2x + 1$
• $v = 6x$

Calculating the derivatives:

• rac{du}{dx} = 24x + 2
• rac{dv}{dx} = 6

Now substituting these values into the quotient rule formula:

$\frac{dy}{dx} = \frac{(6x)(24x + 2) - (12x^2 + 2x + 1)(6)}{(6x)^2}$

Simplifying:

• $= \frac{144x^2 + 12x - 72x^2 - 12x - 6}{36x^2} = \frac{72x^2 - 6}{36x^2} = 2 - \frac{1}{12x^2}$

Thus, rac{dy}{dx} = 2 - \frac{1}{12x^2}.

To find the values of $b$ and $c$, we start by noting that a point of inflection occurs where the second derivative changes sign:

1. First, find the first derivative: $f'(x) = 3x^2 + 2bx + c$

2. Find the second derivative: $f''(x) = 6x + 2b$

Since we have a point of inflection at $x = 2$:

• We set $f''(2) = 0$:
  $6(2) + 2b = 0$

ightarrow 2b = -12 ightarrow b = -6$$ 3. We also know that $f(2) = 4$: $$f(2) = 2^3 + b(2^2) + c(2) - 4 = 4$$ $$8 + 4b + 2c - 4 = 4$$ $$8 - 24 + 2c - 4 = 4$$ $$2c - 20 = 4 ightarrow 2c = 24 ightarrow c = 12$$ Thus, the values of $b$ and $c$ are: $b = -6$ and $c = 12$.