Question 3 (a) Factorise fully: 3xy − 9x + 4y − 12 - Leaving Cert Mathematics - Question 3 - 2019

Starting with the expression for g(x): $g(x) = 3x ext{ln }x - 9x + 4 ext{ln }x - 12$

1. We can reorganize this to: $g(x) = (3x + 4) ext{ln }x - 9x - 12$ Setting this equal to zero gives us: $g(x) = 0$

2. Using the solution from part (a), set: $(3x + 4)( ext{ln }x - 3) = 0$

3. This leads to two cases:

   • Case 1: $3x + 4 = 0$ (not possible since x cannot be negative)
   • Case 2: $ext{ln }x - 3 = 0$

4. Solving for x in case 2 yields:

   $$$$

ightarrow x = e^3$$

Thus, the final solution is: $x = e^3$

To find g′(e), we first differentiate g(x):

1. The derivative g′(x) is determined as follows: g'(x) = 3x imes rac{1}{x} + (3) ext{ln }x - 9 + 4 imes rac{1}{x} = 3 + 3 ext{ln }x - 9 + rac{4}{x}

2. Now, substituting x = e: g'(e) = 3 + 3 ext{ln }e - 9 + rac{4}{e} Since $ext{ln }e = 1$: g'(e) = 3 + 3(1) - 9 + rac{4}{e} = 3 + 3 - 9 + rac{4}{e} = -3 + rac{4}{e}

3. Now, we need to approximate this:

   • Using $e ext{ approximately } 2.71828$, we find: rac{4}{e} ext{ approximately } 1.4715 Thus: $g'(e) ext{ approximately } -3 + 1.4715 = -1.5285$

Finally, rounding to two decimal places gives: $g′(e) ext{ correct to 2 decimal places is } -1.53$