(i) Differentiate the function $f(x) = 4x^3 - 3x^2 + x - 7$, where $x \in \mathbb{R}$, with respect to $x$ - Leaving Cert Mathematics - Question 6 - 2020

To find the slope of the tangent, we evaluate the derivative at the point $x = 1$:

$f'(1) = 12(1)^2 - 6(1) + 1$

Calculating this:

• $f'(1) = 12 - 6 + 1 = 7$.

Therefore, the slope of the tangent at the point $(1, -5)$ is $7$.

We can use the point-slope form of the line to find the equation of the tangent. The point-slope form is given by:

$y - y_1 = m(x - x_1)$

Where $(x_1, y_1)$ is the point $(1, -5)$ and $m$ is the slope ($7$).

Substituting these values in: $y - (-5) = 7(x - 1)$

This simplifies to: $y + 5 = 7x - 7$

Rearranging gives: $7x - y - 12 = 0.$

Thus, the equation of the tangent is: $7x - y - 12 = 0$

Substituting $x = 2$ into $g(x)$ gives: $g(2) = 2(2^2) + p(2) + q = 6$

This simplifies to: $8 + 2p + q = 6$

Rearranging gives: $2p + q = -2 ag{1}$

Next, we evaluate $g'(x)$ for $g'(3)$:

$g'(x) = \frac{d}{dx}(2x^2 + px + q) = 4x + p$

Substituting $x = 3$ gives: $g'(3) = 4(3) + p = 9$

Thus:

ightarrow p = -3 ag{2}$$ Now substituting $p = -3$ back into equation (1): $$2(-3) + q = -2$$ This simplifies to: $$-6 + q = -2 ightarrow q = 4$$ Therefore, the values are: - $p = -3$ - $q = 4$