(a) g(x) = x^3 - 7x^2 + x - 12, ext{ where } x ext{ is in } ext{R} - Leaving Cert Mathematics - Question 4 - 2022

To find the derivative g'(x), we differentiate g(x):

g'(x) = rac{d}{dx}(x^3 - 7x^2 + x - 12) $$= 3x^2 - 14x + 1.$$

Therefore, the derivative is g'(x) = 3x^2 - 14x + 1.

We know that g'(5) = 6. Thus, the slope of the tangent line at x = 5 is 6. Now, we can use the point-slope form of the line:

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

Where (x_1, y_1) = (5, g(5)) = (5, -57) and m = 6:

$$(y + 57) = 6(x - 5)$$

Expanding this:

$$y + 57 = 6x - 30$$

Rearranging into the form ax + by + c = 0:

$$6x - y - 87 = 0.$$

Thus, the equation of the tangent is 6x - y - 87 = 0.

From the graph, u'(x) is negative for any value of x in the interval (0, 2). For example:

x = 1 ext{ (for } 0 < x < 2 ext{)}.$$

On the diagram, the tangent to u(x) at the point (4, 2) approximately has a slope of 1.5 (this can be determined visually from the drawn tangent line). Hence, we estimate:

tu'(4) ext{ is approximately } 1.5.$$