# SIT787 - Mathematics for AI Assignment

- Subject Code :
SIT787

- Country :
Australia

**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.

**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.

In general, if the graph of a function f has a “corner” or “kink” in it, then the graph of thefunction does not have a unique tangent at this point, and f is not differentiable there. Inaddition, 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.

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 xss=removed xss=removed>

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.

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 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.

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

For example

**Question 2. Consider the function**

f(x) = |x _{2} - 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.

**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).

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 = x_{0} and an absolute minimum at x = x_{1}.

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,

(a) Find critical points of f(x), and evaluate f at these points.

(b) 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 xss=removed>
- 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.

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