8.) Let f(z) = z² – 8z + 20

a) Find the discriminant of the quadratic function f
Discriminant= -16

b) Find the complex roots of equation f(z) = 0 in the form z = x ± yi
z= (8±4i)/ 2 = 4±2i

c) Use factorisation to express f in the form of f(z) = (z-h)²+k
(z-4-2i)(z-4-2i)
= (z-4)² + 4

9.) Given the function f(x) = x² – 3bx + (c+2), determine the values of b and c such as f(1) = 0 and f’(3) = 0.

f(1) = 1 – 3b + c + 2 = 0
f’(x) = 2x – 3b
f’(3) = 6 – 3b = 0
b = 2
1 – 3(2) + c + 2 = 0
c =3

10.) If ln(2x-1), find d²y/dx²

dy/dx = 2/2x-1
d²y/dx² = 2(2x-1)^-2(2)
d²y/dx² = -4/(2x-1)²

11.) Find ∫((3x² – x + 2√x)/3√x)dx

= ∫(x^1.5 – ⅓ x^0.5 + ⅔ )dx = (x^2.5/2.5) – (x^1.5/4.5) + (⅔)x + c

Section B

12.) Consider the function f(x) = 2×3 + 3×2 – 12x + 5.

a) Find the stationary points of f(x) and determine their nature (whether they are local maximum, local minimum, or neither).
Solution:
Finding stationary points and their nature:
Step 1: Find the first derivative of f(x)
f'(x) = 6x² + 6x – 12
Step 2: Set f'(x) = 0 to find stationary points
6x² + 6x – 12 = 0
Step 3: Solve for x
Dividing both sides by 6, we get:
x²+ x – 2 = 0
Factoring the quadratic equation, we get:
(x + 2)(x – 1) = 0
Setting each factor to zero, we get two possible values for x:
x + 2 = 0 or x – 1 = 0
Solving for x, we get:
x = -2 or x = 1
So, the stationary points of f(x) are at x = -2 and x = 1.
Step 4: Determine the nature of the stationary points
To determine the nature of the stationary points, we can use the second derivative test. Taking the second derivative of f(x):
f”(x) = 12x + 6
Evaluating f”(x) at x = -2 and x = 1:
f”(-2) = 12(-2) + 6 = -18
f”(1) = 12(1) + 6 = 18
Since f”(-2) is negative, the stationary point at x = -2 is a local maximum. Since f”(1) is positive, the stationary point at x = 1 is a local minimum.

b) Find the equation of the tangent line to the curve of f(x) at the point where x = 2.
Solution:
Finding the equation of the tangent line at x = 2:
Step 1: Find the slope of the tangent line
The slope of the tangent line to the curve of f(x) at any point x can be found using the first derivative f'(x). So, we can evaluate f'(2):
f'(2) = 6(2)² + 6(2) – 12 = 48
So, the slope of the tangent line at x = 2 is 48.
Step 2: Find the point of tangency
The point of tangency is the point where x = 2, so we can plug x = 2 into the original function f(x):
f(2) = 2(2)³ + 3(2)2² – 12(2) + 5 = 37
So, the point of tangency is (2, 37).
Step 3: Write the equation of the tangent line
Using the slope and the point of tangency, we can write the equation of the tangent line in point-slope form:
y – y₁ = m(x – x₁)
Plugging in the values:
y – 37 = 48(x – 2)
Simplifying and rearranging, we get:
y = 48x – 95
So, the equation of the tangent line to the curve of f(x) at the point where x = 2 is y = 48x – 95.

c) Finding the area under the curve of f(x) between x = 0 and x = 4:
Solution:
Step 1: Find the indefinite integral of f(x)
The indefinite integral of f(x) is given by:
∫f(x)dx = ∫(2x³+ 3x² – 12x + 5)dx
Integrating each term separately:
∫2x³dx + ∫3x² dx – ∫12xdx + ∫5dx
= (1/2)x⁴ + x³ – 6x² + 5x + C
where C is the constant of integration.
Step 2: Evaluate the definite integral
To find the area under the curve of f(x) between x = 0 and x = 4, we can evaluate the definite integral:
∫[f(x)]dx from x = 0 to x = 4
= [(1/2)(4)⁴ + (4)³ – 6(4)² + 5(4)] – [(1/2)(0)⁴ + (0)³ – 6(0)² + 5(0)]
= 128 + 64 – 96 + 20
= 116
So, the area under the curve of f(x) between x = 0 and x = 4 is 116 square units.

d) Finding the x-coordinate of the point where the curve of f(x) has a horizontal tangent:
Solution:
Step 1: Find the critical points
The points where the curve of f(x) has a horizontal tangent are the points where the first derivative f'(x) = 0. From part (a), we already found the first derivative f'(x) = 6x² + 6x – 12. Setting f'(x) = 0 and solving for x:
6x² + 6x – 12 = 0
Dividing both sides by 6, we get:
x² + x – 2 = 0
Factoring the quadratic equation, we get:
(x + 2)(x – 1) = 0
Setting each factor to zero, we get two possible values for x:
x + 2 = 0 or x – 1 = 0
Solving for x, we get:
x = -2 or x = 1
So, the critical points of f(x) are at x = -2 and x = 1.
Step 2: Determine the x-coordinate of the point with a horizontal tangent
Since the x-coordinate of the point with a horizontal tangent is the point where the first derivative f'(x) = 0, we can see that x = -2 is the x-coordinate of the point where the curve of f(x) has a horizontal tangent.
So, the x-coordinate of the point where the curve of f(x) has a horizontal tangent is x = -2.

e) Finding the x-values where the curve of f(x) is concave up or concave down:
Solution:
Step 1: Find the second derivative of f(x)
The second derivative of f(x) is given by:
f”(x) = (f'(x))’
From part (a), we already found the first derivative f'(x) = 6x² + 6x – 12. Taking the derivative of f'(x), we get:
f”(x) = (6x² + 6x – 12)’
Using the power rule and sum rule of differentiation, we get:
f”(x) = 12x + 6
Step 2: Determine the concavity of the curve
The concavity of the curve of f(x) is determined by the sign of the second derivative f”(x). If f”(x) > 0, then the curve is concave up. If f”(x) < 0, then the curve is concave down.
Setting f”(x) = 0 and solving for x:
12x + 6 = 0
Dividing both sides by 12, we get:
x + 1/2 = 0
Solving for x, we get:
x = -1/2
So, the curve of f(x) changes concavity at x = -1/2. For x < -1/2, the curve is concave down, and for x > -1/2, the curve is concave up.

f) Finding the inflection points of the curve:
Solution:
Step 1: Determine the x-coordinate of the inflection point
Since the inflection point is the point where the curve changes concavity, we can see from part (e) that the curve changes concavity at x = -1/2. Therefore, the x-coordinate of the inflection point is x = -1/2.
Step 2: Find the y-coordinate of the inflection point
To find the y-coordinate of the inflection point, we can substitute x = -1/2 into the original function f(x):
f(-1/2) = 2(-1/2)³ + 3(-1/2)³ – 6(-1/2) + 5
Simplifying, we get:
f(-1/2) = -1/8 + 3/4 + 3 + 5
f(-1/2) = 25/8
So, the y-coordinate of the inflection point is y = 25/8.
Therefore, the inflection point of the curve of f(x) is at (-1/2, 25/8).