SIT787 Mathematics for AI Assignment Sample

SIT787 Mathematics for AI 

Question 1. Consider the following logarithmic equation.

(i) Find the value(s) of x satisfying in the equation.

(ii) Determine for what values of x the logarithmic expression on the left side of the equation is defined.
Hint: You need to use the rules of logarithms to solve this problem. In general, for the function
y = logb(x) where the base b > 0 and b ̸= 1
• The domain is (0, ∞), and the range is R
• For appropriate values of a, b, c, r, x1, and x2
– logb(x1x2) = logb(x1) + logb(x2)
– logb( x1 ) = logb(x1) − logb(x2)
– logb(xr) = r logb(x)
– logb(1) = 0 and logb(b) = 1
– log (a) = logc(a) which is called the change of base rule.
b logc(b)
– y = logb(x) is an increasing function for b > 1. It means if x1 ≤ x2 then logb(x1) ≤
logb(x2). If 0 < b < 1, the function y = logb(x) is a decreasing function.
– logb(0) is not defined.

Figure 1: (a) y = loge(x) = ln(x) (b) y = log0.5(x)

Piecewise defined functions: Sometimes a function is not differentiable at all points. For example, consider the absolute value function

This function is not differentiable at x = 0. As you can see, the graph of the function changes direction (from decreasing to increasing) abruptly when x = 0.

Figure 2: Absolute value function

In general, if the graph of a function f has a “corner” or “kink” in it, then the graph of the function does not have a unique tangent at this point, and f is not differentiable there. In addition, if a function is not continuous at a point x = a, then f is not differentiable at a. So, at any discontinuity, f fails to be differentiable. A third possibility for a function to be non-differentiable is that the curve has a vertical tangent line at x = a. Let’s see some examples with their plots.

2

(a) f (x) = x 3
1
(b) g(x) = x 3
(c) h(x) = x + 1 if x ≥ 0
sin(x) if x < 0.
Let’s see why these functions are not differentiable.
′ √2
(a) f (x) = x 3 . f (x) = 3 3 x . The derivative function is not defined at x = 0. In Figure 3 (a),
you see that f (x) has a corner at x = 0, and the tangent is vertical. This function is differentiable everywhere except x = 0.
′ 1
(b) g(x) = x 3 . g (x) = 3 √3 x2 . The derivative function is not defined at x = 0. In Figure 3

(b), you see that the tangent is vertical at x = 0. This function is differentiable everywhere except x = 0.

(c) h(x) = x + 1 if x ≥ 0 . This function is not continuous at x = 0 as you can see in sin(x) if x < 0.

Figure 3 (c). How can you check whether the function is not continuous at x = 0? well, if

Figure 3: (a) f (x) (b) g(x) (c) h(x)

you plug x = 0 in the upper and lower rules, you get different values. (0) + 1 ̸= sin(0), or 1 ̸= 0. If a function is not continuous at a point, it is not differentiable there.

Furthermore, if we exclude 0, when x > 0, h(x) = x + 1 . Its derivative for x > 0 is h′(x) = 1. When x < 0, h(x) = sin(x). On this interval, h′(x) = cos(x). Therefore, the derivative of this functions is

You see that the derivative of the function just to the left of x = 0 is −1 and just to the right of x = 0 is +1. The absolute value function is differentiable everywhere except for x = 0. You always need to be careful with the break points. We will learn how to find the derivative of this kind of function soon.

Generally, functions are represented as y = f (x) which consists of a single rule on their domain. However, sometimes functions are defined by different rules on different parts of their domain. Such functions are called piecewise defined functions. For example,

The most famous function of this type is the absolute value function.

They even can have more than two rules on different parts of their domains.

A plot of this function is shown below.

Figure 6: A function with 4 rules

In its general form, a piecewise defined function is represented as

A function of this type is continuous if its constituent functions are continuous on the corresponding intervals and there is no discontinuity at each breakpoint. In other words, f1(x) should be continuous on [a, b), f2(x) should be continuous on [b, c), and f3(x) should be continuous on [c, d). In addition, there should not be a discontinuity on the break points x = b and x = c.

For example the function f (x) in Figure 3 is discontinuous, and the function g(x) in Figure 4 is continuous.
A piecewise function for MBA Assignment Expert is differentiable on a given interval in its domain if the following conditions are satisfied in addition to those for continuity mentioned above:

• its constituent functions are differentiable on the corresponding open intervals,

• at the points where two subintervals touch, the corresponding derivatives of the two neighboring subintervals should match.
For example, for f (x) = 1 − x if x ≤ −1
x2 if x > −1.

its derivative can be found as

• On the open interval x < −1 or (−∞, −1), f (x) = 1 − x, and f′(x) = −1.

• On the open interval x > 1 or (−1, ∞), f (x) = x2, and f′(x) = 2x.

• We visually see that the function is not continuous at x = −1, therefore it is not differentiable there. In other words, if you plug x = −1 in the upper and lower rules, you get different values. In addition, at x = −1, the derivative of the first rule is −1 and of second rule is 2(−1). As −1 ̸= −2 it is not differentiable at x = −1. The derivative of
this function is
f′(x) = −1 if x < −1
2x if x > −1.

We observe that the derivative function is not defined at x = −1.

( 1 − x

if x ≤ −1

x2 if x > −1.

• On the open interval x < −1 or (−∞, −1), g(x) = 1 − x , and g′(x) = −1 .

• On the open interval x > −1 or (−1, ∞), g(x) = x2, and g′(x) = 2x.

• At x = −1, the function is continuous (as 1 − −1 = (−1)2), but using the first rule,
g′(−1) = −1 and using the second rule g′(−1) = 2(−1) = −2. However, −1 ̸= −2. This
function is not differentiable at x = −1. Then, the derivative of this function is

g′(x) =

( −1


if x < −1

2x if x > −1.

There are some piecewise defined functions that are continuous and differentiable everywhere. For example

f (x) =

(x − 1 if x < 1

x2 if x ≥ 1 .
Now, let’s solve the following problem.

Question 2. Consider the function
f (x) = |x2 − 3x + 2| + |x − 3|

(i) Find its derivative using the concept of piecewise-defined functions.

(ii) Also, determine the points when the function is not differentiable, if any exists.
[15 + 10 = 25 marks]

Optimisation: With optimisation, we want to find the maximum and minimum of a function. We have two types of maximum and minimum points. Absolute maximum and absolute minimum, and local minimum and local maximum. Let’s define them properly for a function y = f (x).

• Absolute (or global) maximum: A point x = c is called an absolute maximum of f , if for any value in its domain (x ∈ Dom(f )), f (x) ≤ f (c)

• Absolute (or global) minimum: A point x = c is called an absolute minimum of f , if for any value in its domain (x ∈ Dom(f )), f (x) ≥ f (c)

• Local maximum: A point x = c is called a local maximum of f , if for every value x near
c, f (x) ≤ f (c)

• Local minimum: A point x = c is called a local minimum of f , if for every value x near c, f (x) ≥ f (c).

In the following plot, the function has a local maximum at x = x0, a local minimum at x = x1, and a local maximum at x = x2. The point x = x2 is a global maximum as well. There is no absolute minimum for this function.

However, if we are given a closed interval, we can find the absolute maximum and minimum of the function in that interval. For example, in the following plot, in the closed interval [a, b], the function has an absolute maximum at x = x0 and an absolute minimum at x = x1.

To find all these interesting points of a function, we need to find its critical points. When we find them, then we need to classify each critical point as a global/local maximum/minimum. The critical points for a given function y = f (x) are
the points x = c where f′(c) = 0, or
the points x = c where f′(c) does not exist.
Optimisation problems type 1: Finding absolute maximum and minimum of y = f (x) in a given closed interval [a, b]:

To solve this problem,
Find critical points of f (x), and evaluate f at these points.
Find f (a) and f (b).

The maximum value of items in (a) and (b) is the absolute maximum of f (x) in [a, b], and the minimum value of items in (a) and (b) is the absolute minimum of f (x) in [a, b].

Optimisation problems type 2 (First derivative test): Find local maximum and minimum of y = f (x) using first derivative:
Find all critical points of f (x).
If for a critical point x = c, f′ changes from positive to negative (f changes from increasing to decreasing), x = c is a local maximum point.
If for a critical point x = c, f′ changes from negative to positive (f changes from decreasing to increasing), x = c is a local minimum point.
Optimisation problems type 2 (Second derivative test): Find local maximum and minimum of y = f (x) using second derivative:
Find all critical points of f (x).
For a critical point x = c, if f′(c) = 0 and f′′(c) > 0, x = c is a local minimum.
For a critical point x = c, if f′(c) = 0 and f′′(c) < 0, x = c is a local maximum.
if f′′(c) = 0, the test is inconclusive. It does not give any useful information, and we need to use other techniques to decide the type of stationary point.

Question 3. Consider the following function:

f (x) = √x + √a − 2x

Suppose that the product of the maximum value of f (x) and the minimum value of f (x) is √12. Find the value of a.
Fix a = 4, and plot the function with the following considerations
Find all x- and y-intercepts.

Find all the stationary points and classify them. (you may convert the coordinates of the stationary point(s) into decimal for drawing purposes, however, follow the notes on page 1 of this assignment.)

Determine the intervals for which the function is increasing, and the intervals for which the function is decreasing.
Sketch the function by hand based on the information you gained through steps (ii) to (iv). Label all the important points on the graph of the function.
[20 + 5 + 10 + 10 + 5 = 50 marks

Question 1

The given logarithmic equation is
log_x⁡〖((8-log_5⁡x)/log_5⁡x )^log_3⁡x 〗=1
(i)
For f(x)^(g(x))=1, the solutions are:
f(x)=1 or [g(x)=0 and f(x)≠0] or [f(x)= -1 and g(x) is even]
Now,
log_x⁡((8-log_5⁡x)/log_5⁡x )=1 or log_3⁡〖(x)〗=0 or log_x⁡((8-log_5⁡x)/log_5⁡x )= -1
Now, solving the equation
log_3⁡〖(x)〗=0
Since y=log_b⁡x,x=b^y
So, rewrite the above equation
x=3^0=1
So, we get x=1 as the solution

(ii)

Values of x, for which the logarithmic expression defined, known as the domain of this given logarithmic function.
So, we need to find the domain of
log_x⁡〖((8-log_5⁡x)/log_5⁡x )^log_3⁡x 〗
To define the above expression,
((8-log_5⁡x)/log_5⁡x )^log_3⁡x ≥0
Now, we need to find the set of values of x, for which the expression is defined.
((8-log_5⁡x)/log_5⁡x )≤0
Solving the above equation, we get the value of set as
1<x≤625
Hence, the domain is 1<x≤625 or x ∈(1,625]

Question 2

(i)

To get the derivative of the function f(x), we need to consider the function as a piecewise-defined function, based on the intervals where it changes its behaviour. In this instance, there are two intervals:

When x≤1 or x≥2, we have
f(x)= |x^2-3x+2|+|x-3|
=x^2-3x+2+x-3
=x^2-2x-1
When 1<x<2, we have
f(x)= |x^2-3x+2|+|x-3|
=-x^2+3x-2-x+3
=-x^2+2x+1
Therefore, the derivative is given as
f^' (x)=2x-2, when x≤1 or x≥2
And,
f^' (x)= -2x+2, when 1<x<2

(ii)

We must determine whether the derivative exists at each point in the domain in order to identify the points at which the function is not differentiable. At locations where the derivative is not defined or is discontinuous, the function f(x) is not differentiable.

The derivative is continuous everywhere other than at x = 1 and x = 2, where it alternates between formulas, according to the piecewise definition of f'(x). As a result, for both x = 1 and x = 2, the function f(x) cannot be differentiated.

Question 3

(i)

We must take the derivative of f(x) and make it equal to zero to locate the critical points in order to determine the maximum and minimum values of f(x):
Now,
f(x)= √x+√(a-2x)
So,
f^' (x)=1/(2√x)-1/√(a-2x)
Put f^' (x)=0
1/(2√x)-1/√(a-2x)=0
Solving the above, we get
x=a/6
When a = 4,
x=4/6=2/3
Now, f''(x), we get
-1/(4x^(3/2) )-1/(a-2x)^(3/2)
Substitute, the value of x=2/3, in f''(x), we get
-1/(4(2/3)^(3/2) )-1/(4-2*2/3)^(3/2) <0
At, x=2/3, f^'' (x)<0
So, at x =2/3, the critical point is a local maximum.

(ii)

The plot for the given function, when a = 4

(ii)

From the above graph,

There is no x – intercept and y – intercept = 2

(iii)

There is only one stationary point at x = 2/3 i.e. x = 0.6667

(iv)

From the above graph, the graph is increasing x∈(0,2/3)

And decreasing from x∈(2/3,2).

(v)